黎曼积分与菲氏积分

黎曼积分与菲氏积分
回顾过去，上世纪60年，苏联菲氏微积分教材采用的积分定义是十九世纪高斯弟子黎曼给出的积分定义，国内照搬不误。
    进入本世纪，塔尔斯基弟子Keisler推出内容全面覆盖菲氏微积分教科书的非准微积分教材，简明易懂，堪称世界一流。
  本文明附件是非标准黎曼积分的内容，图形省略。    
袁萌   陈启清   1月26日
附件:Non-standard Limann’s INTEGRATION
.1 THE DEFINITE INTEGRAL
We shall begin our study Of the integral calculus in the same way in which we began with the differential calculus-by asking a question about curves in the plane.
Suppose f is a real function continuous on an interval I and consider the
curve y = f(x). Let a < b where a, bare two points in J, and let the curve be above the
x-axis for x between a and b; that is, f(x) ~ 0. We then ask: What is meant by the
area of the region bounded by the curve y = f(x), the x-axis, and the lines x = a and
x = b? That is, what is meant by the area of the shaded region in Figure 4. 1.1? We call this region the region under the curve y = f(x) between a and b.
y
a b X
Figure 4.1 .1 The Region under a Curve
The simplest possible case is where f is a constant function; that is, the curve
is a horizontal line f(x) = k, where k is a constant and k ~ 0, shown in Figure 4.1.2.
In this case the region under the curve is just a rectangle with height k and width
b - a, so the area is defined as
Area= k•(b- a).
The areas of certain other simple regions, such as triangles, trapezoids, and semicircles,
are given by formulas from plane geometry.
175
176 4 INTEGRATION
y
f(X) = k
0 a b X
area = k(b - a)
Figure 4.1.2
The area under any continuous curve y = f(x) will be given by the definite
integral, which is written
fj(x)dx.
Before plunging into the detailed definition of the integral, we outline the main ideas.
First, the region under the curve is divided into infinitely many vertical
strips of infinitesimal width dx. Next, each vertical strip is replaced by a vertical
rectangle of height f (x ), base dx, and area j (x) dx. The next step is to form the sum
of the areas of all these rectangles, called the infinite Riemann sum (look ahead to
Figures 4.1.3 and 4.1.11). Finally, the integral J~ f(x) dx is defined as the standard
part of the infinite Riemann sum.
The infinite Riemann sum, being a sum of rectangles, has an infinitesimal
error. This error is removed by taking the standard part to form the integral.
It is often difficult to compute an infinite Riemann sum, since it is a sum of
infinitely many infinitesimal rectangles. We shall first study finite Riemann sums,
which can easily be computed on a hand calculator.
Suppose we slice the region under the curve between a and b into thin vertical
strips of equal width. If there are n slices, each slice will have width Llx = (b - a)jn.
The interval [a, b] will be partitioned into n subintervals
[x0 , x 1], [x 1 , x2], ••• , [x11 _ 1 , X 11],
where x0 = a,x 1 =a+ Llx,x2 =a+ 2Llx, .. . ,X11 =b.
The points x0 , x 1 , ... , X 11 are called partition points. On each subinterval [xk _ 1 , xk],
we form the rectangle of height f(xk- d. The kth rectangle will have area
From Figure 4.1.3, we can see that the sum of the areas of all these rectangles will be
fairly close to the area under the curve. This sum is called a Riemann sum and is equal
to
f(x 0 ) Llx + f(x 1) Llx + • • • + /(x,_ 1) Llx.
It is the area of the shaded region in the picture. A convenient way of writing Riemann
sums is the "l:-notation" (l: is the capital Greek letter sigma),
h I f(x) Llx = f(x0 ) Llx + /(x 1) Llx + • • • + /(x11 _ 1) Llx.
4.1 THE DEFINITE INTEGRAL 177
f(x)
x6 x 7 = b X
Figure 4.1.3 The Riemann Sum
The a and b indicate that the first subinterval begins at a and the last subinterval ends
at b.
We can carry out the same process even when the subinterval length ~x does
not divide evenly into the interval length b- a. But then, as Figure 4.1.4 shows, there
will be a remainder left over at the end of the interval [a, b], and the Riemann sum will
have an extra rectangle whose width is this remainder. We let n be the largest integer
such that
a+ n ~x.::::; b,
and we consider the subintervals
[xo, xJl, ... , [xn-1, xn], [xll, b],
where the partition points are
x 0 = a, x 1 = a + ~x, x 2 = a + 2 ~x, ... , x" = a + 11 ~x, b.
f(x)
Figure 4.1.4
178 4 INTEGRATION
X11 will be less than or equal to b but X11 + Llx will be greater than b. Then we define
the Riemann sum to be the sum
b I f(x) Llx = f(x0 ) Llx + f(xd Llx + • • • + /(x11 _ 1) Llx + f(x")(b - X 11).
a
Thus given the function f, the interval [a, b ], and the real number Llx > 0, we have
defined the Riemann sum I~ f(x) Llx. We repeat the definition more concisely.
DEFINITION
Let a < h and let Llx be a posltlve real number. Then the Riemann sum
I~ f(x) Llx is defined as the sum
b I f(x) Llx = f(x 0 ) Llx + f(x 1) Llx + • • • + f(x"- d Llx + f(x")(b - x,J
a
where n is the largest integer such that a + n Llx s b, and
x 0 = a, x 1 = a + Llx, • • •, X 11 = a + n Llx, b
are the partition points.
If X0 = b, the last term f(x")(b - X 11) is zero. The Riemann sum I~ f(x) Llx
is a real function of three variables a, b, and Llx,
D L f(x) Llx = S(a, b, Llx).
The symbol x which appears in the expression is called a dummy variable (or bound
variable), because the value of I~ f(x) Llx does not depend on x. The dummy variable
allows us to use more compact notation, writing f(x) Llx just once instead of writing
f(x0 ) Llx, f(x 1) Llx, f(x 2 ) Llx, and so on.
From Figure 4.1.5 it is plausible that by making Llx smaller we can get the
Riemann sum as close to the area as we wish.
f(x)
a
Figure 4.1.5
--.-.....=-..... ,
I" h I
I
I
I
I
I
I
I
b
I
I
I
I
I
I
I
I
I
I
I
X
EXAMPLE 1 Letf(x) = !x. In Figure 4.1.6, the region under the curve from x = 0
to x = 2 is a triangle with base 2 and height 1, so its area should be
A= !bh = 1.
y
Figure 4.1.6
q., I I n t:. LJ C r- I 1'\l I I t:. 11'\l I t:. \.::1 riP\ L
I
f(x) = 2x
Area= l
2 X
Let us compare this value for the area with some Riemann sums. In Figure
4.1.7, we take Llx = l The interval [0, 2] divides into four subintervals
[0, -!-J, [-!-, 1], [1, Hand[!, 2]. We make a table of values ofj(x) at the lower
endpoints.
y
Figure 4.1.7
The Riemann sum is then
2
I t.x = 2
Riemann sum= i
X
"L. .. f( X ) LAl X = 0 ' 21 + 41 '21 + 21 " 21 + 43 '21 = 86 •
0
In Figure 4.1.8, we take Llx = i. The table of values is as follows.
The Riemann sum is
2
I o * 0 t
2 3 4 5 6 7
4 4 4 4 4 4
2 .1 4 5 6 7
8 8 8 8 8 8
" f( ) A 0 1 1 1 2 1 3 1 4 I 5 1 6 1 7 I 7 1... x LlX= '4+8•4+8•4+8'4+8'4+8'4+8'4+8'4=8•
0
We see that the value is getting closer to one.
Finally, let us take a value of Llx that does not divide evenly into the interval
length 2. Let Llx = 0.6. We see in Figure 4.1.9 that the interval then divides
into three subintervals of length 0.6 and one of length 0.2, namely [0, 0.6],
[0.6, 1.2], [1.2, 1.8], [1.8, 2.0].
0 0.6 1.2 1.8
0 0.3 0.6 0.9
""'
y I
2.x = 4
Riemann sum = i
y .h = 0.6
Riemann sum = . 72
0 2 X X
Figure 4.1 .8 Figure 4.1.9
The Riemann sum is
2 L f(x) Llx = 0(.6) + (.3)(.6) + (.6)(.6) + (.9)(.2) = .72.
0
EXAMPLE 2 Letj(x) = j1 - x 2
, defined on the closed interval I= [ -1, 1]. The
region under the curve is a semicircle of radius 1. We know from plane
geometry that the area is n/2, or approximately 3.14/2 = 1.57. Let us compute
the values of some Riemann sums for this function to see how close they are
to 1.57. First take t.x =~as in Figure 4.1.10(a). We make a table of values.
xk - 1 -1/2 0 1/2
f(xk) 0 J3;4 1 J3;4
The Riemann sum is then
I I f(x) Llx = 0 • 1/2 + vl3f4 • 1/2 + 1 • 1/2 + J3;4 • 1/2
-I
= 1 +2J3 ~ 1.37.
Next we take Llx = t. Then the interval [ -1, 1] is divided into ten subintervals
as in Figure 4.1.10(b). Our table of values is as follows.
4 3 2 1 2 3 4
xk -1 0 - - - -
5 5 5 5 5 5 5 5
f(xk)
3 4 '\/21 J24 ~ fo 4 3
0 - - - -
5 5 5 5 5 5 5 5
f(x) f(x)
X X
(a) (b)
4.1 THE DEFINITE INTEGRAL 181
The Riemann sum is
I1 1[ 3 4 fi1 J24 J24 fi1 4 3] f(x) 8.x = - 0 + - + - + - + - + 1 + - + - + - + -
-1 5 5 5 5 5 5 5 5 5
= 19 + 2j21 + 2)24 ~ 1 52.
25 .
Thus we are getting closer to the actual area rr/2 ~ 1.57.
By taking ~x small we can get the Riemann sum to be as close to the area
as we wish.
Our next step is to take ~x to be infinitely small and have an irifinite Riemann
sum. How can we do this? We observe that if the real numbers a and bare held fixed,
then the Riemann sum
b I f(x) 8.x = S(8.x)
a
is a real function of the single variable 8.x. (The symbol x which appears in the
expression is a dummy variable, and the value of
b I f(x) 8.x
a
depends only on 8.x and not on x.) Furthermore, the term
b I f(x) 8.x = S(8.x)
is defined for all real 8.x > 0. Therefore by the Transfer Principle,
b I f(x) dx = S(dx)
a
is defined for all hyperreal dx > 0. When dx > 0 is infinitesimal, there are infinitely
many subintervals of length dx, and we call
b I f(x) dx
a
an infinite Riemann sum (Figure 4.1.11).
f(x)
a X b X
Fi ure 4.1.11 Infinite Riemann um
182 4 INTEGRATION
We may think intuitively of the Riemann sum
b
If(x)dx
as the infinite sum
f(x 0)dx + f(xddx + • • • + f(xH- 1)dx + f(xH)(b- xH)
where H is the greatest hyperinteger such that a + H dx :::;; b. (Hyperintegers are
discussed in Section 3.8.) H is positive infinite, and there are H + 2 partition points
x 0 , x 1, ... , xH, b. A typical term in this sum is the infinitely small quantity f(xx) dx
where K is a hyperinteger, 0 :::;; K < H, and xx =a + K dx.
The infinite Riemann sum is a hyperreal number. We would next like to take
the standard part of it. But first we must show that it is a finite hyperreal number and
thus has a standard part.
THEOREM 1
Let f be a continuous function on an interval I, let a < b be two points in I, and
let dx be a positive infinitesimal. Then the infinite Riemann sum
b I f(x) dx
a
is a finite hyperrealnumber.
PROOF Let B be a real number greater than the maximum value off on [a, b].
Consider first a real number i".x > 0. We can see from Figure 4.1.12 that the
1--------b -a _____ _,
1-------no .6x ----~
Figure 4.1.12 a b
finite Riemann sum is less than the rectangular area B • (b - a);
b I f(x) i".x < B • (b - a).
Therefore by the Transfer Principle,
b I f(x) dx < B • (b - a).
In a similar way we let C be less than the minimum off on [a, b] and show
that
4.1 THE DEFINITE INTEGRAL 183
b L f(x) dx > C • (b - a).
a
Thus the Riemann sum L~ f(x) dx is finite.
We are now ready to define the central concept of this chapter, the definite
integral. Recall that the derivative was defined as the standard part of the quotient
!1yjl1x and was written dyjdx. The "definite integral" will be defined as the standard
part of the infinite Riemann sum
b L f(x) dx,
a
and is written J! f(x) dx. Thus the /1x is changed to dx in analogy with our differential
notation. The ~ is changed to the long thin S, i.e., J, to remind us that the integral is
obtained from an infinite sum. We now state the definition carefully.
DEFINITION
Let f be a continuous function on an interval I and let a < b be two points in I.
Let dx be a positive infinitesimal. Then the definite integral off from a to b with
respect to dx is defined to be the standard part of the infinite Riemann sum with
respect to dx, in symbols
ff(x)dx = st(tf(x)dx).
We also define ff(x)dx = 0,
ff(x) dx = - ff(x) dx.
By this definition, for each positive infinitesimal dx the definite integral
is a real function of two variables defined for all pairs (u, w) of elements of I. The
symbol x is a dummy variable since the value of
does not depend on x.
In the notation L~ f(x) dx for the Riemann sum and J: f(x) dx for the
integral, we always use matching symbols for the infinitesimal dx and the dummy
variable x. Thus when there are two or more variables we can tell which one is the
dummy variable in an integral. For example, x 2 t can be integrated from 0 to 1 with
respect to either x or t. With respect to x,
1 L x 2t dx = x~t dx + xft dx + • • • + x]i_ 1t dx
0
184 4 INTEGRATION
(where dx = 1/H), and we shall see later that f x 2t dx = st(x6t dx + :.;it dx + • • • + x}1 _ 1t dx) = 1t.
With respect to t, however,
I L x 2t dt = x 2 t0 dt + x 2 t1 dt + • • • + x 2
tK-J dt,
0
and we shall see later that
11 xzt dt = ixz.
The next two examples evaluate the simplest definite integrals. These
examples do it the hard way. A much better method will be developed in Section 4.2.
EXAMPLE 3 Given a constant c > 0, evaluate the integral g c dx.
y
Figure 4.1.13 shows that for every positive real number L1x, the finite Riemann
sum is
b L c L1x = c(b - a).
By the Transfer Principle, the infinite Riemann sum in Figure 4.1.14 has the
same value,
b L c dx = c(b - a).
Taking standard parts,
fcdx = c(b- a).
ll
This is the familiar formula for the area of a rectangle.
1+--fl il.x--
c ~•--•--r---,,---.---~~
a Xn b X a X b X
Figure4.1.13 Figure 4.1.14
4.1 THE DEFINITE INTEGRAL 185
EXAMPLE 4 Given b > 0, evaluate the integral Jt x dx.
(1)
(2)
(3)
(4)
The area under the line y = x is divided into vertical strips of width dx.
Study Figure 4.1.15. The area of the lower region A is the infinite Riemann
sum
b
area of A = L x dx.
0
By symmetry, the upper region B has the same area as A;
area of A = area of B.
Call the remaining region C, formed by the infinitesimal squares along the
diagonal. Thus
area of A +area of B +area of C = b2
•
Each square in C has height dx except the last one, which may be smaller,
and the widths add up to b, so
0 :::.;; area of C :::.;; b dx.
Putting (1)-(4) together,
2 t x dx :::.;; b2
:::.;; ( 2 t x dx) + b dx.
Since b dx is infinitesimal,
b
2l_:X dx::::::: b2
,
0
b b2
Ixdx ::::::: 2 .
0
Taking standard parts, we have
rb b2
Jo xdx = 2•
B
Figure 4.1.15
186 4 INTEGRATION
PROBLEMS FOR SECTION 4.1
Compute the following finite Riemann sums. If a hand calculator is available, the Riemann sums
can also be computed with L>.x = ftJ.
3
5
7
9
11
13
15
17
19
21
23
D 25
D 26
D 27
I~ (3x + 1) L>.x, L>.x = 1 2 L,~ (3x + 1) L>.x,
I~ 1 (3x+1)L>.x, L>.x = i 4 I~ 2x2 L>.x, L>.x = i
I~ 1 2x2 L>.x, L>.x = i 6 Ig (2x - 1) L>.x, L>.x = 1
Ig (2x - 1) L>.x, L>.x = 2 8 I~t (xz- 1)L>.x, L>.x =!
I~ (x2
- 1) L>.x, L>.x =! 10 I~t (x2- 1)L>.x, L>.x =-fa
I~ 4 (5x 2 - 12)L>.x, L>.x = 2 12 I~ 4 (5x 2 - 12) L>.x, L>.x = 1
Ii (1 + 1/x) L>.x, L>.x = 1 14 Is 10-h L>.x 0 •. L>.x =!
I~~ x4 L>.x, L>.x = i 16 I~ 1 2x3 L>.x, L>.x =!
I~ .fi L>.x, L>.x = 1 18 I~2lx- 41 L>.x, L>.x = 2
I~ sinx L>.x, L>.x = rr4 20 I" sin2
0 -' L>.-x- . L>.x = rr,4
I~ e' L>.x. L>.x = 1 5 22 '[,~ xe' L>.x, L>.x =I 5
L, '- In' L>.' L>.x = 1 24 I'- -ln-x L>.x I - -' I X .. L>.x = 1
Let b be a positive real number and ll a positive integer. Prove that if L>.x = bjn,
b
I x L>.x = (1 + 2 + • • • + (n - 1)) l>.x 2
.
0
_ . ll(n- 1)
Usmg the formula 1 + 2 + • • • + (11- 1) = --
2
-, prove that
b I x L>.x = (1 - 1/nW/2.
0
Let H be a positive infinite hyperinteger and dx = bjH. Using the Transfer Principle and
Problem 25, prove that Jt x dx = b2 /2.
Let b be a positive real number, n a positive integer, and L>.x = bjn. Using the formula
2 22 32 ( 1
)2 n(n - 1)(2n - 1)
1 + + +"•+ n- =
6
,
prove that
~ .z A _ n(n - 1)(2n - 1) b3
L. X LlX -
6 3 .
0 ll
D 28 Use Problem 27 to show that Jt x 2 dx = b3 /3.
4.2 FUNDAMENTAL THEOREM OF CALCULUS
In this section we shall state five basic theorems about the integral, culminating in
the Fundamental Theorem of Calculus. Right now we can only approximate a
definite integral by the laborious computation of a finite Riemann sum. At the end
of this section we will be in a position easily to compute exact values for many definite
integrals. The key to the method is the Fundamental Theorem. Our first theorem
shows that we are free to choose any positive infinitesimal we wish for dx in the
definite integral.
4.2 FUNDAMENTAL THEOREM OF CALCULUS 187
THEOREM 1
Given a continuous function f on [a, b] and two positive infinitesimals dx
and du, the definite integrals with respect to dx and du are the same,
f f(x) dx = f!(u) du.
From now on when we write a definite integral J~ f(x) dx, it is understood
that dx is a positive infinitesimal. By Theorem 1, it doesn't matter which infinitesimal.
The proof of Theorem 1 is based on the following intuitive idea. Figure 4.2.1
shows the two Riemann sums I~ f(x) dx and I~ f(u) du. We see from the figure
that the difference I~ f(x) dx - I~ f(u) du is a sum of rectangles of infinitesimal
height. These difference rectangles all lie between the horizontal Jines y = -E and
y = E, where E is the largest height. Thus -E(b -a) s I~ f(x) dx - I~ f(u) du s
e(b - a). Taking standard parts,
0 s f f(x) dx - f f(u) du S: 0,
f f(x) dx = f f(u) du.
j(x)
b
y = -e
Figure 4.2.1
188 4 INTEGRATION
Theorem 1 shows that whenever Ll.x is positive infinitesimal, the Riemann
sum is infinitely close to the definite integral,
Ib fb f(x) Ll.x ~ f(x) dx.
a a
This fact can also be expressed in terms of limits. It shows that the Riemann sum
approaches the definite integral as Ll.x approaches 0 from above, in symbols
f f(x) dx =8-~~~+ t f(x) Ll.x.
Given a continuous function f on an interval I, Theorem 1 shows that the
definite integral is a real function of two variables a and b,
A(a, b) = f f(x) dx, a, bin I.
We now formally define the area as the definite integral shown m Figure 4.2.2.
a b X
Figure 4.2.2
DEFINITION
Iff is continuous and f(x) ;::-:.: 0 on [a, b], the area of the region below the
curve y = f(x)ji•om a to b is defined as the definite integral:
Area = f f(x) dx.
The next two theorems give basic properties of the integral.
THEOREM 2 (The Rectangle Property)
Suppose f is continuous and has minimum value m and maximum value M
on a closed interval [a, b]. Then
m(b - a) s f f(x) dx s M(b - a).
That is, the area of the region under the curve is between the area of the rectangle
whose height is the minimum value off and the area of the rectangle \\'hose
height is the maximum value off in the interval [a, b].
4.2 FUNDAMENTAL THEOREM OF CALCULUS HS~
The Extreme Value Theorem is needed to show that the minimum value m
and maximum value M exist. The rectangle of height miscalled the inscribed rectangle
of the region, and the rectangle of height M is called the circumscribed rectangle.
From Figure 4.2.3, we see that the inscribed rectangle is a subset of the region under
the curve, which is in turn a subset of the circumscribed rectangle. The Rectangle
Property says that the area of the region is between the areas of the inscribed and
circumscribed rectangles.
y
M
m
a b X
Figure 4.2.3 The Rectangle Property
PROOF By Theorem 1, any positive infinitesimal may be chosen for dx. Let us
choose a positive infinite hyperinteger H and let dx = (b - a)/H. Then
dx evenly divides b - a; that is, the interval [a, b] is divided into H subintervals
of exactly the same length dx. Then
b I m dx = m • H • dx = m(b - a),
a
b I M dx = M • H • dx = M(b - a).
For each x, we have m ~f(x) ~ M. Adding up and taking standard parts,
we obtain the required formula.
b b b I m dx ~I f(x) dx ~I M dx,
a a
m(b- a)~ f f(x)dx ~ M(b- a).
One useful consequence of the Rectangle Property is that the integral of
a positive function is positive and the integral of a negative function is negative:
Ifj(x) > 0 on [a, b], then 0 < m(b - a) ~ f f(x) dx.
Ifj(x) < 0 on [a, b], then f f(x) dx ~ M(b - a) < 0.
The definite integral of a negative function f(x) = - g(x) from a to b is
just the negative of the area of the region above the curve and below the x axis.
This is because
f(x) dx = -g(x) dx,
(See Figure 4.2.4.)
a
Figure 4.2.4
b b I f(x) dx = - I g(x) dx,
a a
ff(x) dx = - f g(x) dx.
1------...... -... .........
........... g(.r)
.......
+ ~---'~
b
THEOREM 3 (The Addition Property)
Supposefis continuous on an interml I. Then for all a, b, c in I,
f f(x) dx = f f(x) dx + f f(x) dx.
This property is illustrated in Figure 4.2.5 for the case a < b < c. The
Addition Property holds even if the points a, b, c are in some other order on the
real line, such as c < a < b.
f(x)
Figure 4.2.5 a b c
PROOF First suppose that a < b < c. Choose a dx that evenly divides the first
interval length b - a. This simplifies our computation because it makes
b a partition point, b = a + H dx. Then, as Figure 4.2.6 suggests,
c b c I f(x) dx = I f(x) dx + I f(x) dx.
b
Taking standard parts we have the desired formula
J>(x) dx = f f(x) dx + f f(x) dx.
4.2 FUNDAMENTAL THEOREM OF CALCULUS 191
Figure 4.2.6 a b c
To illustrate the other cases, we prove the Addition Property when
c < a < b. The previous case gives
f f(x) dx = r f(x) dx + f f(x) dx.
Since reversing the endpoints changes the sign of the integral,
- J:j(x)dx =- {f(x)dx + fj(x)dx,
and the desired formula
f f(x) dx = f j(x) dx + f f(x) dx
follows.
The definite integral of a curve can be thought of as area even if the curve
crosses the x-axis. The curve in Figure 4.2.7 is positive from a to band negative from
b to c, crossing the x-axis at b. The integral f~ j(x) dx is a positive number and the
integral Sb j(x) dx is a negative number. By the Addition Property, the integral
{f(x)dx = f j(x)dx + J:f(x)dx
is equal to the area from a to b minus the area from b to c. The definite integral
s~ f(x) dx always gives the net area between the x-axis and the curve, counting
areas above the x-axis as positive and areas below the x-axis as negative.
The definite integral f~ f(t) dt is a real function of two variables u and v
and does not depend on the dummy variable t. If we replace u by a constant a and v
by the variable x, we obtain a real function of one variable x, given by
F(x) = r j(t) dt.
Our fourth theorem states that this new function is continuous.
f(x)
e x
Figure 4.2.7
192 4 INTEGRATION
THEOREM 4
Let j be continuous on an intenal I. Choose a point a in I. Then the fimction
F(x) defined by
F(x) = rj(t) dt
is continuous on I.
SKETCH OF PROOF Let c be in I, and let x be infinitely close to c and between
the endpoints of I. By the Addition Property, r j(t) dt = rf(t) dt + J>(t) dt,
Jc f(t) dt - J' j(t) dt = f f(t) dt,
a a x
and F(c) - F(x) = L f(t) dt.
This is the area of the infinitely thin strip under the curve y = f(t) between
t = x and t = c (see Figure 4.2.8). The strip has width ~x = c - x. By the
Rectangle Property, its area is between m ~x and M ~x and hence is infinitely
small. Therefore F(x) is infinitely close to F(c), and F is continuous on I.
f(t)
/F(c) -F(x)
Figure 4.2.8 a c b
Our fifth theorem, the Fundamental Theorem of Calculus, shows •that
the definite integral can be evaluated by means of antiderivatives. The process of
antidifferentiation is just the opposite of differentiation. To keep things simple, let I
be an open interval, and assume that all functions mentioned have domain I.
DEFINITION
Let j and F be functions with domain I. Iff is the derivative ofF, then F is
called an antiderivative off.
4.2 FUNDAMENTAL THEOREM OF CALCULUS 193
For example, suppose a particle is moving upward along the y-axis with
velocity v = f(t) and position y = F(t) at time t. The position y = F(t) is an antiderivative
of the velocity v = f(t). We shall discuss antiderivatives in more detail
in the next section. We are now ready for the Fundamental Theorem.
FUNDAMENTAL THEOREM OF CALCULUS
Suppose f is continuous on its domain, which is an open interval I.
(i) For each point a in I, the definite integral offfi"om a to x considered as a
function ofx is an antiderivative off That is,
d(f f(t) dt) = f(x) dx.
(ii) IfF is any antideriuative of j; then for any two points (a, b) in I the
definite integral off fi"om a to b is equal to the difference F(b) - F(a), r f(x) dx = F(b) - F(a).
The Fundamental Theorem of Calculus is important for two reasons. First,
it shows the relation between the two main notions of calculus: the derivative, which
corresponds to velocity, and the integral, which corresponds to area. It shows that
differentiation and integration are "inverse" processes. Second, it gives a simple
method for computing many definite integrals.
EXAMPLE 1
(a) Find f~ c dx. Since ex is an antiderivative of c, f c dx = cb - ca = c(b - a).
(b) Find f.b x dx. ix2 is an antiderivative of x. Thus
ll
The above example gives the same result that we got before but is much
simpler. We can easily go further.
EXAMPLE 2 Find J~ x2 dx. x 3j3 is an antiderivative of x2 because
d(x 3/3) 3x2
---=-=x.2
dx 3
Therefore fb b3 a3
a xzdx = 3-3'
This gives the area of the region under the curve y = x2 between a and b
(Figure 4.2.9).
194 4 INTEGRATION
y
y = x2
a b X
Figure 4.2.9
If a particle moves along the y-axis with continuous velocity u = f(t), the
position y = F(t) is an antiderivative of the velocity, because v = dyjdt. The
Fundamental Theorem of Calculus shows that the distance moved (the change in y)
between times t = a and t = b is equal to the definite integral of the velocity,
distance moved = F(b) - F(a) = f f(t) dt.
EXAMPLE 3 A particle moves along they-axis with velocity v = 8t 3 cmjsec. How
far does it move between times t = -1 and t = 2 sec? The function
G(t) = 2t 4 is an antiderivative of the velocity v = 8t 3
. Thus the definite
integral is
distance moved = J2
8t3 dt = 2 o 24
- 2 o ( -1)4 = 30 em.
-[
EXAMPLE 4 Find J6 jt dt (Figure 4.2.10). The function yft is defined and continuous
on the half-open interval [0, oc ). But to apply the Fundamental
Theorem we need a function continuous on an open interval that contains
the limit points 0 and 4. We therefore define
f(t) = {)
fort < 0
fort ::0: 0.
This function is continuous on the whole real line. In particular it is continuous
at 0 because if t ~ 0 then f(t) ~ 0. The function
Figure 4.2.10
fort < 0
for t ::0: 0
4.2 FUNDAMENTAL THEOREM OF CALCULUS 195
is an antiderivative of f. Then f jt dt = F(4) - F(O) = (t. 43/2 - t. o3t2) = lt
In the next section we shall develop some methods for finding antiderivatives.
The antiderivative of a very simple function may turn out to be a "new" function
which we have not yet given a name.
EXAMPLE 5 The only way we can show that the functionf(x) =~has an
antiderivative is to take a definite integral
fJi+7dt.
This is a "new" function that cannot be expressed in terms of algebraic,
trigonometric, and exponential functions without calculus.
The Fundamental Theorem can also be used to find the derivative of a
function which is defined as a definite integral with a variable limit of integration.
This can be done without actually evaluating the integral.
EXAMPLE 6 Let y = f.J1+t2 dt. Then y = -r .j1+t2 dt,
X 2
and
J
xl+x 1
EXAMPLE 7 Let y = -3--dt.
3 t + 1
Let u = x2 + x. Then
du - = (2x + 1),
dx
By the Chain Rule,
Ju 1
y = -3--dt,
3 t + 1
dy
du u3 + 1•
dy = dy du = _1_(lx + 1) = 2x + 1 .
dx du dx u3 + 1 (x2 + x)3 + 1
We conclude this section with a proof of the Fundamental Theorem of
Calculus.
PROOF (i) Let F(x) be the area under the curve y = f(t) from a to x,
F(x) = r f(t) dt.
Imagine that the vertical line cutting the t-axis at x moves to the right as
in Figure 4.2.11.
196 4 INTEGRATION
a X
Figure 4.2.11
We show that the rate of change of F(x) is equal to the length f(x) of the
moving vertical line.
Suppose x increases by an infinitesimal amount ,1.x > 0. Then
F(x + ,1.x) - F(x) = r+t.x f(t) dt
is the area of an infinitely thin strip of width ,1.x and height infinitely close to
f(x). By the Rectangle Property the area of the strip is between the inscribed
and circumscribed rectangles (Figure 4.2.12),
m ,1.x :::;; F(x + ,1.x) - F(x) :::;; M ,1.x.
Dividing by ,1.x,
F(x + ,1.x) - F(x)
m :::;; ,1.x :::;; M.
Since f is continuous at x, the values m and Mare both infinitely close to f(x),
and therefore
F(x + ,1.x) - F(x) ""' f( )
A ""' X.
uX
The proof is similar when ,1.x < 0. Hence F'(x) = f(x).
a
Figure 4.2.12
X
F(x+ 6x)- F(x)
t:.x
4.2 FUNDAMENTAL THEOREM OF CALCULUS 197
PROOF (ii) Let F(x) be any antiderivative off Then, by (i),
d(F(x)- rf(t)dt) =f(x)- f(x) = 0.
In Section 3.7 on curve sketching, we saw that every function with derivative
zero is constant. Thus
F(x)- ff(t) dt = C0 , F(x) = f'f(t)dt + C0
for some constant C0 . Then
so
F(b) - F(a) = (f f(t) dt + C0 ) - (f f(t) dt + C0 )
= f f(t) dt - 0 = f f(t) dt,
F(b) - F(a) = f f(x) dx.
PROBLEMS FOR SECTION 4.2
In Problems 1-14, find an antiderivative of the given function.
1 f(x) = sfi 2 f(x) = 4/fi
3 f(t) = 3t2 + 1 4 f(x) = 5x3
5 f(t) = 4- 3t2 6 f(z) = 2/z2
7 f(s) = 7s-3 8 f(t) = t2 + t-2
9 f(x) = (x - 6)2 10 f(u) = (Su + 1)2
11 f(y) = y3f2 12 f(x) = 2/xfi
13 f(x) = lxl 14 f(t) = 12t - 41
15 If F'(x) = x + x 2 for all x, find F(1) - F( -1).
16 If F'(x) = x4 for all x, find F(2) - F(1).
17 If F'(t) = t 113 for all t, find F(8) - F(O).
Evaluate the definite integrals in Problems 18-22.
18 J1
2x2 dx 19 Iz x3 dx
-1 -2
20 I-I t-2dt 21 f 2fidx
-2
22 r2 -5x4 dx
-3
In Problems 23-27 an object moves along they-axis. Given the velocity v, find how far the object
moves between the given times t0 and t 1 .
23 v = 2t + 5, t 0 = 0, t 1 = 2
24 v = 4- t, t0 = 1, t 1 = 4
198 4 INTEGRATION
25
26
27
l' = 3,
t' = 3t2
,
r = 1or- 2 ,
t0 = 2, It = 6
t0 = 1, It = 3
10 = 1, t t = 100
In Problems 28-32, find the area of the region under the curve y = f(x) from a to b .
28
29
30
31
32
33
34
D 35
D 36
D 37
D 38
D 39
.\' = 4- X
2
,
y = Jx + 2,
.r = 9x- x2
•
\, = ;-:;:- X
. 'v" . '
.\' = 3xt 3
a= -2, b = 2
a= -2. b = 2
a= 0, b = 3
(/ = 0, b = 1
a= 1, b = 8
If F'(t) = t - 1 for all t and F(O) = 2, find F(2).
If F'(x) = 1 - x 2 for all x and F(3) = 5, find F(- 1).
Suppose F(x) and G(x) have continuous derivatives and F'(x) + G'(x) = 0 for all x.
Prove that F(x) + G(x) is constant.
Suppose F(x) and G(x) have continuous derivatives such that F'(x) :<; G'(x) for all x.
Prove that F(b) - F(a) :<; G(b) - G(a)
where a <b.
Prove that a function F(x) has a constant derivative if and only if F(x) is linear, i.e., of the
form F(x) = ax + b.
Prove that a function F(x) has a constant second derivative if and only if F(x) has the
form F(x) = ax2+ hx + c.
Suppose that F"(x) = G"(x) for all x. Prove that F(x) and G(x) differ by a linear function,
that is, G(x) = F(x) + ax + b for some real numbers a and b.
4.3 INDEFINITE INTEGRALS
The Fundamental Theorem of Calculus shows that every continuous function f
has at least one antiderivative, namely F(x) = J~ f(t) dt. Actually, f has infinitely
many antiderivatives, but any two antiderivatives off differ only by a constant. This
is an important fact about antiderivatives, which we state as a theorem.
THEOREM 1
Let f be a real function whose domain is an open interval I.
(i) If F(x) is an antiderivative of f(x), then F(x) + C is an antiderivative
off(x)for every rea/number C.
(ii) If F(x) and G(x) are two antiderivatives of f(x), then F(x) - G(x) is
constant for all x in I. That is,
G(x) = F(x) + C
for some real number C.
Discussion Parts (i) and (ii) together show that if we can find one antiderivative
F(x) off(x), then the family of functions
F(x) + C, C = a real number
4.3 INDEFINITE INTEGRALS 199
gives all antiderivatives of f(x). We see from Figure 4.3.1 that the graph
of F(x) + Cis just the graph of F(x) moved vertically by a distance C. The
graphs of F(x) and F(x) + C have the same slopes at every point x. For
example, letj(x) = 3x2
• Then F(x) = x 3 is an antiderivative of 3x2 because
d(x3)- 3 2
~-X.
But x3 + 6 and x3 - J2 are also antiderivatives of 3x2. In fact, x3 + Cis
an antiderivative of 3x2 for each real number C. Theorem 1 shows that 3x2
has no other antiderivatives.
y
X
Figure 4.3.1
PROOF We prove (i) by differentiating,
d(F(x) +C)= d(F(x)) dC =f( ) O =f( )
dx dx + dx x + x •
Part (ii) follows from a theorem in Section 3.7 on curve sketching. If a
function has derivative zero on I, then the function is constant on I. The
difference F(x) - G(x) has derivative f(x) - f(x) = 0 and is therefore
constant. We used this fact in the proof of the Fundamental Theorem of
Calculus.
In computing integrals of J, we usually work with the family of all antiderivatives
off We shall call this whole family of functions the indefinite integral off
The symbol for the indefinite integral is J f(x) dx. If F(x) is one antiderivative of J,
the indefinite integral is the set of all functions of the form F(x) + C0 , C0 constant.
We express this with the equation
f f(x) dx = F(x) + C.
It is an equation between two families of functions rather than between two single
functions. C is called the constant of integration. To illustrate the notation,
J 3x2 dx = x3 + C.
We repeat the above definitions in concise form.
200 4 INTEGRATION
DEFINITION
Let the domain off be an open interval I and suppose f has an antiderivative.
The family of all antiderivatives off is called the indefinite integral off and is
denoted by f f(x) dx.
Given a function F, the family of all functions which differ fi'om F only by a
constant is written F(x) + C. Thus ifF is an antiderivative off we write
J f(x) dx = F(x) + C.
When working with indefinite integrals, it is convenient to use differentials
and dependent variables. If we introduce the dependent variable u by u = F(x), then
du = F'(x) dx = f(x) dx.
Thus the equation Jf(x) dx = F(x) + C
can be written in the form J du = u +C.
The differential symbol d and the indefinite integral symbol J behave as
inverses to each other. We can start with the family of functions u + C, form du, and
then form J du = u + C to get back where we started. Some of the rules for differentiation
given in Chapter 2 can be turned around to give a set of rules for indefinite
integration.
THEOREM 2
Let u and v be functions of x whose domains are an open interval I and suppose
du and dv exist for every x in I.
(i) J du = u + C.
(ii) Constant Rule f c du = c f du.
(iii) Sum Rule f du + dv = J du + f dv.
(iv) Power Rule f ur+l
u' du = --+ C, where r is rational, r =F - 1, }' + 1
and u > 0 on I.
(v) J sinudu = -cosu +C.
(vi) J cos u du = sin u + C.
(vii) J e" du = e" + C.
4.3 INDEFINITE INTEGRALS 201
(viii) J ~du =In lui+ C (u =f. 0).
Discussion The Power Rule gives the integral of u' when r =f. - 1, while Rule
(viii) gives the integral of u' when r = -1. When we put u = f(x) and
v = g(x), the Constant and Sum Rules take the form
Constant Rule f cf(x) dx = c f f(x) dx.
Sum Rule f (f(x) + g(x)) dx = f f(x) dx + f g(x) dx.
In the Constant and Sum Rules we are multiplying a family of functions
by a constant and adding two families of functions. If we do either of these
two things to families of functions differing only by a constant, we get another
family of functions differing only by a constant. For example,
7(3x4 + C) = 2lx4 + 7C = 21x4 + C'
is the family of all functions equal to 2lx4 plus a constant. Similarly,
(3Jx +C)+ (5x- Jx +D)= 5x + 2Jx + (C +D)= 5x + 2Jx + C'
is the family of all functions equal to 5x + 2Jx plus a constant.
PROOF OF THEOREM 2
(i) This is just a short form of the theorem that u + C is the family of all
functions which have the same derivative as u.
(ii) We have c du = d(cu), whence
f c du = f d(cu) = cu + C = c(u + C') = c f du.
(iii) du + dv = d(u + v),
f du + dv = f d(u + v) = u + v + C = f du + f dv.
(iv) d ( -- = du = u du,
u' + 1
) (r + 1 )u' r
r+l r+l
f ur+l
u' du = --
1
+ C.
r+
Rules (v)-(viii) are similar. Only the last formula, (viii), requires an explanation.
The absolute value in In I u I comes about by combining the two cases u > 0
and u < 0. When u > 0, u = I u I and
1
d(ln lui)= d(ln u) = -du.
u
When u < 0, In u is undefined, but lui= -u and In lui= In ( -u). Thus
1 1
d(ln lui)= d(ln ( -u)) = - -d( -u) = -du.
u u
202 4 INTEGRATION
Thus, in both cases, when 11 =1= 0,
1
d(ln lui)= -du,
II
J ~ du = In lui + C.
u
EXAMPLE 1 J (2x- 1 + 3 sinx)dx = 2ln lxl- 3 cosx +C.
We can use the rules to write down at once the indefinite integral of any
polynomial.
EXAMPLE 2 J (4x3 - 6x2 + 2x + 1) dx = x4
- 2x3 + x2 + X + C.
3 2
EXAMPLE 3 -~+3X3/2+C.
Indefinite integration is much harder than differentiation, because there are
no rules for integrating the product or quotient of two functions. It often requires
guesswork. The short list of rules in Theorem 1 will help, and as this course proceeds
we shall add many more techniques for finding indefinite integrals.
EXAMPLE 4 Show that ( dx J1 +X J (1 + x)1i2(1 - x)3!2 = 1 - x + C.
Our rules give no hint on finding this integral. However, once the answer
is given to us we can easily prove that it is correct by differentiating,
dj1 +X
1 _ x d((1 + x)1;2(1 _ x)-112)
dx dx
= (1 + x)1'2(- 1)(-!)(1 _ x)-3;2 + (1 _ x)-1;2(!)(1 + x)-1 12
= (1 + x)- 1
' 2(1- x)- 3
:
2 [!(1 + x) + ±0- x)]
1
Here is a warning that may prevent some common mistakes.
Warning: The integral of the product of two functions is not equal to the
product of the integrals. The same goes for quotients. That is,
Wrong: J (uv) dx = (f u dx) (f v dx).
4.3 INDEFINITE INTEGRALS 203
For example,
Wrong:
Correct: f f x3 xz
x(x + 1) dx = (x2 + x) dx = ~ + 2 +C.
Wrong:
f u Ju dx -dx=-.
v f vdx
For example,
Wrong: f (x + 1) dx 1 2 f X + 1 d = = {l•)x + X C Jx x f Jx dx (~)x312 +
= 3.._/~ + _3_ + c.
4 2Jx
Correct: f xJx
1
dx = f(Jx + Jx) dx = ~x312 + 2Jx +c.
The indefinite integral can be used to solve problems of the following type.
Given that a particle moves along the y-axis with velocity v = f(t), and that at a
certain timet = t0 its position is y = y0 . Find the position y as a function oft.
EXAMPLE 5 A particle moves with velocity v = 1jt2
, t > 0. At time t = 2 it is at
position y = 1. Find the position y as a function oft. We compute
f v dt = f _!_ dt = - ~ + c.
t2 t
Since dyjdt = v, y is one of the functions in the family -1/t + C. We can
find the constant C by setting t = 2 andy = 1,
1
y =-- + C,
t
Then the answer is
1
1 = - l + C,
1 1 y = -- + lz.
t
c = 1!.
The next theorem shows that in such a problem we can always find the answer
if we are given the position of the particle at just one point of time.
<!U4 4 INTEGRATION
THEOREM 3
Suppose the domain off is an open interred I and f has an antideriratire. Let
P(x0 , y0 ) be any point with x 0 in I. Then f has exactly one antideriratiL•e
whose graph passes through P.
PROOF Let F be any antiderivative off Then F(x) + C is the family of all antiderivatives.
We show that there is exactly one value of C such that the
function F(x) + C passes through P(x0 , y0 ) {Figure 4.3.2). We note that all
of the following statements are equivalent:
( 1) F(x) + C passes through P(x0 , y0 ).
(2) F(x0 ) + C = Yo•
(3) C = Yo - F(x0 ).
Thus y0 - F(x0 ) is the unique value of C which works.
y
X
Figure 4.3.2
The Fundamental Theorem of Calculus, part (ii), may be expressed briefly
as follows, where f is continuous on I.
Iff f(x) dx = F(x) + C, then f f(x) dx = F(b)- F(a).
For evaluating definite integrals we introduce the convenient notation
J
b
F(x) " = F(b) - F(a).
It is read "F(x) evaluated from a to b."
The Constant and Sum Rules hold for definite as well as indefinite integrals:
Constant Rule
ib (b cf(x) dx = c J" f(x) dx.
Sum Rule fucx) + g(x)) dx = f f(x) dx + f g(x) dx.
The Constant Rule is shown by the computation
4.3 INDEFINITE INTEGRALS 205
f cf(x) dx = cF(b) - cF(a) = c(F(b) - F(a)) = c f f(x) dx.
The Sum Rule is similar.
EXAMPLE 6 Evaluate the definite integral of y = (1 + t)jt 3 from t = 1 to t = 2
(see Figure 4.3.3).
f2 1 + t f2 -3-dt = (t- 3 + t- 2)dt
1 t 1
= f t- 3 dt+ t- 2 dt=- +- 2 J2 C2]2 t-1]2
1 1 -2 1 -1 1
( 1 1 ) (1 1) 3 1 7 = ( -2). 4- ( -2) •1 + -2- -=1 = 8 + 2 = 8'
Thus the area under the curve y = (1 + t)jt3 from t = 1 to t = 2 is l
y
Figure 4.3.3
y=!+-t
t"
2
EXAMPLE 7 Find the area of the region under one arch of the curve y = sin x
(see Figure 4.3.4).
One arch of the sine curve is between x = 0 and x = n. The area is the
definite integral
fo" sin x dx = -cos x J:
= -cos n- (-cos 0) = -( -1) - ( -1) = 2.
The area is exactly 2.
y
Y= sin x
X
Figure 4.3.4
206 4 INTEGRATION
EXAMPLE 8 Find the area under the curve y = - 2x- 1 from x = -5 to x = -1.
(See Figure 4.3.5.)
The area is given by the definite integral
J-1
~ 5 -2x- 1 dx.
First compute the indefinite integral
J -2x- 1 dx = -2 J x- 1 dx = -2ln lxl +C.
Now compute the definite integral.
J-1
-2x- 1 dx = -2ln lxl]-
1
-5 -5
= - 2(ln I - 11 - In I -51) = - 2(ln 1 - In 5)
= 2ln 5 ~ 3.219.
y
y = -2x-•
-5 -I X
Figure 4.3.5
This example illustrates the need for the absolute value in the integration rule
J x- 1 dx =In lx I+ C.
The natural logarithm In x is undefined at x = -5 and x = -1, but In I xI is defined
for all x =1= 0. The absolute value sign is put in when integrating x- 1 and removed
when differentiating In I x 1.
EXAMPLE 9 In computing definite integrals one must first make sure that the
function to be integrated is continuous on the interval. For instance,
IncmTect: J1
-;dx = x-
1
]
1
= -1- (-(-1)) = -2.
_ 1 x -1 -1
This is clearly wrong because 1/x2 > 0 so the area under the curve cannot be
negative. The mistake is that 1/x2 is undefined at x = 0 and hence the
function is discontinuous at x = 0. Therefore the area under the curve and
the definite integral
are undefined (Figure 4.3.6).
f(x)
f(x) = _J_
x2
Figure 4.3.6
PROBLEMS FOR SECTION 4.3
Evaluate the following integrals.
J (1 + 2x + 3x2)dx
3 f (12t7
- 3t5 + 2t2 + 1) dt
5 f (t!f2 + t-lf2)dt
7 J (2x- 3)2 dx
9 J (z + 1/z)2 dz
11 J 5 cos x dx
13 J x + 1 dx
X
15 J (1 + x- 1
)
2 dx
17 J (3 + jt)(4 - 2jt) dt
19 f4 + 3JY + yJY
y2
dy
21 J (ax2 + bx + c)dx
4.3 INDEFINITE INTEGRALS 207
Jl 1
2 dx
-IX
X
2
4
6
8
10
12
14
16
18
20
22
~F(x)
-!
F(x) = ~
-I
J (2x2
- 6x + 9)dx
f (5 + y-2- 4y-3)dy
f (2ii3 - 3y2i3) dy
J (x- 2)(2x + 1)dx
J (z - 1/z)2 dz
J (sinx + cosx)dx
J 2x 2
- 3x + 6
xz
dx
J 3ex dx
J 3s + 1 ds
3Js
J (3 - x2)(1 + 4x2
) dx
J (a 3x3 + a2x
2 + a 1x + a0) dx
208 4 INTEGRATION
23 J2
(2x - 4x 3 + x 5
) dx 24 11
(I + x 2 + 3x4) dx
-2
25 r (I + x 2 + 3x4
) dx 26 r e-'dx
-I -I
27 1" cos x dx 28 r2
0
cos x dx
J1
2
29 3x-1d x 30 f -X--d1 x
2 X
31 r11- dx
-3 X
In Problems 32-36, find the position y as a function oft given the velocity v = dyjdt and the value
of)' at one point of time.
32 v = 2t + 3, y = 0 when t = 0
33 v = 4t 2
- 1, r-"> when t=O
34 l' = 3r4
, y=O when t= -l
35 r = 2 sin r, J' = 10 when t=O
36 r = 31- 1, .r=1 when t = 1
In Problems 37-42, find the position y and velocity vas a function oft given the acceleration a and
the values of y and v at 1 = 0 or t = 1.
37
38
39
40
41
42
43
44
45
a= I, l' = 0 and )' = 1 when I= 0
a= -32, l' = 10 and y=O when t=O
a= 31 2
, V=1 and )'=2 when t = 0
-
a= 1 - .Jt, r = -2 and )' = 1 when 1=0
a= t- 3 , r = 1 and .r=O when t = 1
a= -sin t, r = 0 and )'=4 when 1=0
Which of the following definite integrals are undefined? f1 f1 (a) -dx (b) - dx
-I X 1-'
(c) r i~dr (d) r ' /\-dr .. ,. - .
-I -I (e) r/ 4- x2 dx (f) f2' x 2
- 4 dx
-2
(g) r -,-I du
-lu--1
(h) rl _
2
, t 2
- I d1
(i) f2 v~ldt (JJ r lx- 11 dx
-3
(k) J1
tan x dx (I) 1" tan xdx
-I
Find the function I such that I' is constant. f(O) = f'(O) and /(2) = f'(2).
An object moves with acceleration a = 6t. Find its position y as a function oft. given
that y = I when t = 0 and .r = 4 when t = I.
46
0 47
Find the function h such that h" is constant, h(l) = I, h(2) = 2. and h(3) = 3.
Suppose that F"(x) exists for all x. and let (x 0 , y0 ) and (x 1 , y 1) be two given points.
Prove that there is exactly one function G(x) such that
4.4 INTEGRATION BY CHANGE OF VARIABLES 209
G(xo) =Yo
G'(x 1) = y 1
G"(x) = F"(x) for all x.
l 48 Assume that F"(x) exists for all x, and let (x 1 , y 1) and (x 2 , y2 ) be two points with x 1 =f= x 2 .
Prove that there is exactly one function G(x) such that G"(x) = F"(x) for all x, and the
graph of G passes through the two points (x 1 , Ytl and (x2 , y2 ).
~ INTEGRATION BY CHANGE OF VARIABLES
We have seen that the sum, constant, and power rules for differentiation can be turned
around to give the sum, constant, and power rules for integration. In this section we
shall show how to make use of the Chain Rule for differentiation in problems of
integration. The Chain Rule will lead to the important method of integration by
change of variables. The basic idea is to try to simplify the function to be integrated
by changing from one independent variable to another.
IfF is an antiderivative off and we take u as the independent variable, then
f f(u) du is a family of functions of u,
J f(u) du = F(u) + C.
But if we take x as the independent variable and introduce u as a dependent variable
u = g(x), then du and f f(u) du mean the following:
du = g'(x) dx, f f(u) du = f f(g(x))g'(x) dx = H(x) + C.
The notation f f(u) du always stands for a family of functions of the independent
variable, which in some cases is another variable such as x. The next theorem can be
used as follows. To integrate a given function of x, properly choose a new variable
u = g(x) and integrate a new function with respect to u.
DEFINITION
Let I and J be intervals. We say that a function g maps J into I if for every
point x in J, g(x) is defined and belongs to I (Figure 4.4.1 ).
y
I
J X
Figure 4.4.1 g maps J into I
.!IU 4 INII:::.llHAIIUN
THEOREM 1 (Indefinite Integration by Change of Variables)
Suppose I and J are open intervals, f has domain I, g maps J into I, and g is
differentiable on J. Assume that when we take u as the independent variable,
Jf (u) dti = F(u) + C.
Then when x is the independent variable and u = g(x),
J f(u) du = F(g(x)) + C.
PROOF Let H(x) = F(g(x)). For any x in J, the derivatives g'(x) and F'(g(x)) = f(g(x))
exist. Therefore by the Chain Rule,
H'(x) = F'(g(x))g'(x) = f(g(x))g'(x).
It follows that
J f(g(x))g'(x) dx = H(x) + C = F(g(x)) -\- C.
So when u = g(x), we have
f(u) du = f(g(x))g'(x) dx, J f(u) du = F(g(x)) + C.
Theorem 1 gives another proof of the general power rule
J
un+!
u" du = --
1
+ C.
n+
n =1- -1,
where u is given as a function of the independent variable x, from the simpler power
rule
J x"+t
x"dx = -- + C,
11 + 1
n =1- -1,
where x is the independent variable.
EXAMPLE 1 Find f(4x + 1)3 + (4x + 1j2 + (4x + 1) dx. Let u = 4x + 1. Then
du = 4 dx, dx = ± du. Hence
J (4x + 1)3 + (4x + 1)2 + (4x + 1)dx
= J (u3 + u2 + u), ~ du = ~(u4 + u3 + uz) + C
4 4 4 3 2
_ ~ [(4x + 1)4 (4x + 1)3 (4x + 1)2
]
-4 4 + 3 + 2 +C.
EXAMPLE 2 Find J 2( -
1
I 2 dx.
x I + 1 x)
Let u = I + 1/x. Then du = -1/x2 dx and thus
So
4.4 INTEGRATION BY CHANGE OF VARIABLES 211
1 ---+c.
1 + 1/x
In a simple problem such as this example, we can save writing by using the
term 1 + 1/x instead of introducing a new letter u,
f -1 dx = f 1 d( 1 + ~) = (1 + 1/x)-' + C
x2(1 + 1/x)2 (1 + 1/xf x - 1 •
In examples such as the above one, the trick is to find a new variable u such
that the expression becomes simpler when we change variables. This usually must
be done by an "educated" trial and error process.
One must be careful to express dx in terms of du before integrating with
respect to u.
EXAMPLE 3 Find f(l + 5xf dx. Let u = 1 + 5x. For emphasis we shall do it
correctly and incorrectly.
Correct: du = 5 dx, dx = t du,
f f 1 u3 (1 + 5x)3
(1 + 5x)2 dx = u
2
• -5
du = - + C = + C.
. 15 15
Incorrect: f f u3 (1 + 5x)3
(1 + 5xf dx = u2 dx = 3 + C =
3
+ C.
Incorrect: f f u3 (1 + 5x) 3
(1 + 5x)2 dx = u2 du = 3 + C =
3
~ + C.
EXAMPLE 4 Find J x3 j2- x2 dx. Let u = 2 - x2
, du = - 2x dx, dx = du/(- 2x).
We try to express the integral in terms of u.
J x3 j2 - x2 dx = J x\ru :~x = J - ~ x\ru du.
Since u = 2 - x2
, x2 = 2 - u. Therefore
J -tx2 Ju du = J -!(2- u)Ju du = J -Ju + !u 3
i
2 du
-~u3f2 + t, ~ustz + c
= -~2 _ x2)3/2 + t(2 _ xz)stz + C.
We next describe the method of definite integration by change of variables. In
a definite integral f h(x) dx
it is always understood that x is the independent variable and we are integrating
between the limits x = a and x = b. Thus when we change to a new independent
LlL 4 INTEGRATION
variable u, we must also change the limits of integration. The theorem below will
show that if u = c when x = a and u = d when x = b, then c and d will be the new
limits of integration.
THEOREM 2 (Definite Integration by Change of Variables)
Suppose I and J are open intervals, f is continuous and has an antiderivative
on I, g has a continuous derivative on J, and g maps J into I. Then for any two
points a and b in J.
fb fg(b)
f(g(x))g'(x) dx = f(u) du.
a g(a)
PROOF Let F be an antiderivative of f Then by Theorem 1, H(x) = F(g(x)) is an
antiderivative of h(x) = f(g(x))g'(x). Since f, g, and g' are continuous, h is
continuous on J. Then by the Fundamental Theorem of Calculus,
fb f~~ f(g(x))g'(x) dx = H(b) - H(a) = F(g(b)) - F(g(a)) = f(u) du.
a g(a)
EXAMPLE 5 Find the area under the line y = 1 + 3x from x = 0 to x = 1. This can
be done either with or without a change of variables.
Without change of variable: f(1 + 3x) dx = x + 3x2 /2 + C, so f (1 + 3x) dx = x + 3~2I = ( 1 + 3 ~12) - ( 0 + 3 ~oz) ~
With change of variable: Let u = 1 + 3x. Then du = 3 dx, dx = 1 du.
When x = 0, 11 = 1 + 3 • 0 = 1. When x = 1, u = 1 + 3 • 1 = 4.
J1 J4 1 l/2]4 (1 + 3x) dx = u •- du = -
0 1 3 6 1
16
6 6
15 5
6 2'
Example 5 shows us that J6 (1 + 3x) dx = Ji (u/3) du; that 1s, the areas
shown in Figure 4.4.2 are the same.
y v
y = 1 + 3x
X II
Figure 4.4.2
y
2
2x y=-(
x2-3)2
3
4.4 INTEGRATION BY CHANGE OF VARIABLES 213
v
X
I v=u2
6 u
Figure 4.4.3
EXAMPLE 6 Find the area under the curve y = 2xj(x2
- 3)2 from x = 2 to x = 3
(Figure 4.4.3).
Let u = x2 - 3. Then du = 2x dx. At x = 2, u = 22 - 3 = 1. At x = 3,
u = 32
- 3 = 6. Then
J3 2x dx= (6_.!._du=-~]6=1-~=~.
2 (x2
- 3)2 J 1 u2 u 1 6 6
EXAMPLE 7 Find g~ x dx. The function~ x as given is only defined
on the closed interval [ -1, 1]. In order to use Theorem 2, we extend it to the
open interval J = (- oo, oo) by
h(x) = {J1- x2x
if x < - 1 or x > 1,
if -1:::;: X:::;: 1.
Let u = 1 - x2
. Then du = -2x dx, dx = -duj2x. At x = 0, u = 1. At
x = 1, u = 0. Therefore
f ~xdx = f Ju•C-±du) = f- ±Judu
= ± f Ju du = ± •1u312 ]~ = 1 - 0 = 1.
We see in Figure 4.4.4 that as x increases from 0 to 1, u decreases from 1 to 0,
so the limits become reversed. The areas shown in Figure 4.4.5 are equal.
u
X
Figure 4.4.4 u = 1 -x2
214 4 INTEGRATION
y v
.fJI-x2 xdx
u
X ro _l Vii du
}I 2
Figure 4.4.5
We can use integration by change of variables to derive the formula for the
area of a circle, A = nr2
, where r is the radius. It is easier to work with a semicircle
because the semicircle of radius r is just the region under the curve
-r :s; x :s; r.
To start with we need to give a rigorous definition of n. By definition, n is the area of
a unit circle. Thus n is twice the area of the unit semicircle, which means:
DEFINITION
n=2J1 ~dx.
-I
The area of a semicircle of radius r is the definite integral
J~, vlr2
- x 2 dx.
To evaluate this integral we let x = ru. Then dx = r du. When x = ±r, u = ± 1. Thus
f, .Jr2
- x2 dx = f 1
Jr2
- (ru) 2 r du = f 1
r\/~~~ du
= r2 J1 ~du = r2 •?'_.
-1 2
Therefore the semicircle has area nr2 /2 and the circle area nr2 (Figure 4.4.6).
f1 3x2
- 1
EXAMPLE 8 Find ~ dx.
o1+y~..-.-..-.
Let u = x - x 3
. Then du = (1 - 3x2
) dx. When x = 0, u = 0 - 03 = 0.
When x = 1, tl = 1 - 1 3 = 0. Then
f 1 3x2
- 1 fo du
---===dX = - = 0
o 1 + Jx- x 3 o 1 + Ju .
As x goes from 0 to 1, u starts at 0, increases for a time, then drops back to 0
(Figure 4.4.7).
4.4 INT~GKATION BY CHANGE 01- VARIAI:!LE~
y v
-r r -1 u
-rrr2 Jl ~ 2= _1 r 2vl-u2 du
Figure 4.4.6
f(x) ll
u = x- x 3
X X
Figure 4.4.7
We do not know how to find the indefinite integrals in this example. Nevertheless
the answer is 0 because on changing variables both limits of integration
become the same. Using the Addition Property, we can also see that. for
instance,
f2!3 3x2 - 1 il 3x2 - 1
----=== dx = - dx.
0 1+~ 2;31+~
PROBLEMS FOR SECTION 4.4
In Problems 1-90, evaluate the integral.
1 f (2x ~ 1)2 dx
2 f J3Y+1 dy
3 f (3- 4z)6 dz 4 f (1 - x)3i2 dx
5 f2t~dt 6 f X d
J2x2 + 1 x
7 f x(4 + 5x2)2 dx 8 f 4y
(2 + 3y2)2 dy
~10
~ 11\IIC.\...Jn/-\J IUJ\1
9 J sin(3x) dx 10 J cos(4 - 2x) dx
11 J 6sin(4x -1)dx 12 J asinx + bcosxdx
13 J sinO cosO dB 14 J sin2 8 cos 8 dB
15 J cos3 8 sin 8 dB 16 J sin (28) + cos (38) dB
17 J x sin(x2 + 1) dx 18 J x2 cos(x3
) dx
19 J sin (In x) dx 20 J e' cos(e') dt
X
21 J fiiiU cost dt 22 J "/i cos(tjt) dt
23 J e2•' dx 24 J 3e1 -x dx
25 J ae•' + be-x dx 26 J (e"' + 1)2 dx
27 J xe"'' dx 28 J xe1
-•'' dx
29 J be""' dx 30 J eax+b dx
31 J esin 8 cos 8 dB 32 Je'~dt
33 J-
1
-dx
x+2
34 J 3 _: 4xdx
35 J --e"'d x 36 J~dx
e"' + 1 X +
37 j--' dx
X+ 1
38 J x2 (1 ~ 1/x) dx
39
J vh(1 ~ ~)dx 40 J 1 - 2t dt
1 + 2t
41 J 3t + 4 dt 42 J x2 32dx
5- t X +
43 J x3)X4+5dx 44 J 1 dx
(2x - 1 )j 1 - 2x •
45 J )' /l+?dy . ' .
46 J (2t2 ~ 1)3 dt
47 J --1-1 du 48 J (4x + 1)j2x2 + x + 5dx
v fl-U2
49 J 1
j3s + 2
ds 50 J jl=Szdz
51 J 1 dx 52 J 1 dy
x2J1 + 1/x /(1 - 4/y)2
53 Jx- 3~-2 dx 54 J 1 ~ dx
jx(l + 2Jx)2
4.4 INTEGRATION BY CHANGE OF VARIABLES 217
55 r3- ~)2 d• Jx X
56 JJ1)t jt dt
57 I 2 + 1/z dz
zz 58 I (3y +1 1)3 dy
59 f X 2 d
jx3 + 4 x
60 I xz
~
dx
61 f x3 dx 62 Jx 3y'1=7dx J1+7
63 I tjt+l dt 64 Ls: 2)3 ds
65 J (2s + 6)(1 - s)- 4 ds 66 f y3j4+Yldy
67 I )'3
(yl + 1)3 dy 68 f xs
J1+7 dx
69 f x dx
J4X+I
70 J J2 + .,fo du
71 Iu~dr1 72 I 1
(2~+W
dx
73 I 4x- 1 dx 74 J- -x-z dx
J4X+I fx-=-1
75 f x3
1- x4dx 76 f y3
2- yl dy
77 I ys
1 + yl dy 78 I (u; 4)2 du
79 f 6!1- 5
(3u + 2f du 80 f- --dx 1
1+~
81 I ~ d 82 I!?+ c?sx dx
2+~ X ex+ SlllX
83 J c?se de
sme
84 J tane de
85 I --1 dx
a+ bx
86 I 2x + 1 d
X2 +X+ 1 X
87 I sine de
1 +case
88 I s~ne- case de
sme + cose
89 Il:x dx 90 I --1 dx
xlnx
In Problems 91-108, evaluate the definite integral.
91 f/3
0 sine de 92
fn/4
- n/4 cos (2e) de
93 II ex dx 94 11
xi?, dx
-I
r1 96 r 95 -dx -2X- -1 dx
1 2x 0 X +
97 f/2
0 sine cose de 98 rn 0 a sine + b cos e de
218
99
101
103
105
107
109
llO
111
112
113
114
D 115
D 116
D 117
D liS
D 119
4 INTEGRATION
fF+idx 100 r 1 o (4x + W dx
L4 (2x + 1 )Ji2 dx 102 Ll t(t 2 + 3)- 2 dt
f (1 + 6x-j3 dx 104 r 2
I ;,!3t+l dt
fv~dv 106 Jl xz
-I (4 - x3)2 dx r~ Xd x 108 Is x(xz + 2)1i3 dx
-1 X J6
Find the area of the region below the curve y = 1/(10 - 3x) from x = 1 to x = 2.
Find the area of the region under one arch of the curve y = sin x cos x.
Find the area of the region under one arch of the curve y = cos (3x).
Find the area of the region below the curve y = 4xj4=Xl between x = 0 and x = 2.
Find the area below the curve y = (1 + 7x)213 between x = 0 and x = 1.
Find the area below the curve y = x/(x2 + 1)2 between x = 0 and x = 3.
J1 1- 2x
Evaluate:
3r:: dx
o 1 + J x- x2
Evaluate: f 1
2xj(1 - x2
)
3 + 1 dx
Let f and g have continuous derivatives and evaluate J f'(g(x))g'(x) dx.
A real function f is said to be even if f(x) = f(- x) for all x. Show that iff is a continuous
even function, then J~J(x) dx = J~ f(x) dx.
An odd function is a real function g such that g(- x) = - g(x) for all x. Prove that for
a continuous odd function g, s~b g(x) dx = 0.
4.5 AREA BETWEEN TWO CURVES
A region in the plane can often be represented as the region between two curves.
For example, the unit circle is the region between the curves
y = -~, y = jl - x 2
, -1 :o;; x :o;; 1
shown in Figure 4.5.1. Consider two continuous functions f and g on [a, b] such that
f(x) :::; g(x) for all x in [a, b]. The region R, bounded by the curves
y = f(x), y = g(x), x = a, x = b,
is called the region betweenf(x) and g(x) from a to b. If both curves are above the
x-axis as in Figure 4.5.2, the area of the region R can be found by subtracting the
area below f from the area below g:
area of R = f g(x) dx - f f(x) dx.
It is usually easier to work with a single integral and write
area of R = f(g(x) - f(x)) dx.
4.5 AREA BETWEEN TWO CURVES 219
y
y
y=~
X
f(x)
a b X
y = -Vl=X2
Figure 4.5.1 Figure 4.5.2
In the general case shown in Figure 4.5.3, we may move the region R above the
x-axis by adding a constant c to bothf(x) and g(x) without changing the area, and
the same formula holds:
y
a
Figure 4.5.3
area of R = f(g(x) +c) dx- f(f(x) +c) dx
= f(g(x)- f(x))dx.
y
X a
f(x)
I
I
1 f(x) + c
I g(x)
I
I
I
I
f(x)
I
I
I
I
I
I
I il
b
To sum up, we define the area between two curves as follows.
X
220 4 INTEGRATION
DEFINITION
Iff and g are continuous and f(x) ::::: g(x) for a ::::; x ::::; b, then the area of the
region R between f(x) and g(x) fi'om a to b is defined as f (g(x) - f(x)) dx.
EXAMPLE 1 Find the area of the region between the curves y = !x2
- 1 andy = x
from x = 1 to x = 2. In Figure 4.5.4, we sketch the curves to check that ±x2
- 1 ::::; x for 1 ::::; x ::::; 2. Then
A = fx- C±x2
- 1) dx = ±x2
- ix3 +xi=~-
y
X
Figure 4.5.4
EXAMPLE 2 Find the area of the region bounded above by y = x + 2 and below
by y = x 2
.
Part of the problem is to find the limits of integration. First draw a sketch
(Figure 4.5.5). The curves intersect at two points, which can be found by
solving the equation x + 2 = x 2 for x.
x 2
- (x + 2) = 0, (x + 1)(x - 2) = 0,
x = - I and x = 2.
Then A = J2
(x + 2 - x2
) dx = !x2 + 2x - tx 3
]
2
= 4±.
-1 -1
-1 2 X
Figure 4.5.5
4.5 AREA BETWEEN TWO CURVES 221
EXAMPLE 3 Find the area of the region R bounded below by the line y = -1 and
above by the curves y = x3 and y = 2 - x. The region is shown in Figure
4.5.6.
y
X
(3, -I)
Figure 4.5.6
This problem can be solved in three ways. Each solution illustrates a different
trick which is useful in other area problems. The three corners of the region
are:
( -1, -1),
(3, -1),
(1, 1),
where y = x3 and y = - 1 cross.
where y = 2 - x and y = - 1 cross.
where y = x3 and y = 2 - x cross.
Note that y = x3 and y = 2 - x can cross at only one point because x3 is
always increasing and 2 - xis always decreasing.
FIRST SOLUTION Break the region into the two parts shown in Figure 4.5.7:
R1 from x = -1 to x = 1, and R2 from x = 1 to x = 3. Then
Figure 4.5.7
area of R = area of R1 + area of R2 .
area of R1 = f
1
x3
- (-1)dx = !x4 + xi
1
= 2.
area of R2 = f (2 - x) - ( -1) dx = 3x -!x2I = 2.
area of R = 2 + 2 = 4.
y
X
First solution
ZZZ 4 INTEGRATION
SECOND SOLUTION Form the triangular regionS between y = -1 andy= 2- x
from -1 to 3. The region R is obtained by subtracting from S the regionS 1
shown in Figure 4.5.8. Then
area of R = area of S- area of S1 .
area of S = f3
(2 - x) - ( -1) dx = 3x - !x2]~
-I I
= 8.
area of S1 = J1
(2- x)- x3 dx = 2x- !x2
- ±x4l 1
= 4.
-I J -I
area of R = 8 - 4 = 4.
y
X
(3, -I)
Figure 4.5.8 Second solution
THIRD SOLUTION Use y as the independent variable and x as the dependent
variable. Write the boundary curves with x as a function of y.
y=2-x
y = x3
becomes x = 2 - y.
becomes x = yl 13 .
The limits of integration are y = -1 and y = 1 (see Figure 4.5.9). Then
A = Jl (2- y)- y1!3 ely= 2)'- ty2 - ;iy4i3]' = 4.
-1 -I
As expected, all three solutions gave the same answer.
X
Fi ure 4.5.9 Tlmd solution
Lf-.0 1-\nC.M DC I YYI.-1.-1'11 I YV...., -..;....., 11 v ._....,
PROBLEMS FOR SECTION 4.5
In Problems 1-43 below, sketch the given curves and find the area of the region bounded by
them.
1 f(x) = 0, g(x) = 5x - x2
, 0 ::; x ::; 4
2 f(x) = Jx, g(x) = x2
, 1 ::; x ::; 4
3 f(x) = x~, g(x) = 1, -1 ::; x ::; 1
4 y = X - 2, y = 3x113, 0 ::; X ::; 1
5 }' = Jx, }' = Jx+"l, 0 ::; X ::; 4
6 y = P+1-X, }' = P+1 + X, - 0::; X ::; 1
7 The x-axis and the curve y = - 5 + 6x - x2
8 The x-axis and the curve y = 1 - x4
9 The y-axis and the curve x = 25 - y2
10 The y-axis and the curve x = y(S - y)
11 }' = COS X, }' = 2 COS X, - nj2 ::; X ::; nj2
12 y = sin x cos x, y = 1, 0::; x::; n/2
13 y = -sin x, y = sin x, 0 ::; x ::; n
14 y = sin x, y = cos x, 0 ::; x ::; n/4
15 y = sin x cos x, y = sin x, 0 ::; x ::; n
16 y = sin 2 x cos x, y = sin x cos x, 0 ::; x ::; n/2
17 y = x, y = ex, 0 ::; x ::; 2
18 y = e-x, y = eX, 0 ::; x ::; 2
19 }' = -ex, }' = ex, - 1 ::; X ::; 1
20 y = xex>, y = e, 0 ::; x ::; 1
21
22
23
24
1
}' = X + 1, }' = 1, 0 ::; X ::; 2
1 1
}' = 2x + 1 ' y = X + 1 ' O ::; X ::; 2
}' = 1/x, }' = X, 1 ::; X ::; 2
X
y = xz + 1 , y = t, 0 ::; x ::; 1
25 f(x) = x3i2, g(x) = x2i3
26 y = x 2 - 2x, y = x - 2
27 y = x4
- 2x2
, y = 2x2 + 12
28 y = x4 - 1, y = x3 - x
29 y = x4 /(x2 + 1), y = 1/(x2 + 1)
30 }'=X 3~, }'=X~, O:s;x
31 y = 2x2
, y = x 2 + 4
32 X = y2, X = 2 - y2
33 Jx + JY = 1 and the x- and y-axes
34 x 2y = 4, x2 + y = 5 (first quadrant)
35 y = xJx+"l, y = 2x
36 }' = 0, y = x3 + X + 2, X = 2
37 y = 2x + 4, y = 2 - 3x, y = - x
38 v=x2 -l. v=(x-1f, y=(x+1f
;!;!4
39
40
41
42
43
44
45
46
47
48
49
50
51
52
4 INTI=liRATIUN
)' = ,/x, )' = L y = 10- 2x
)' = X - 2, )' = 2 - X, )' = j_~
J' = -X, _\' = p, )' = 3x - 2
J' = -2, J' = x 3 + X, X + J' = 3
r = x 2 , )' = 2x- 2
, _\' = 2x- 3 (first quadrant)
Find the area of the ellipse x 2 fa 2 + y2/b 2 = I. Use the fact that the unit circle has area n.
Sketch the four-sided region bounded by the lines y = 1, y = x, y = 2x, and
y = 6 - x and find its area.
Find the number c > 0 such that the region bounded by the curves y = x, y = - 2x, and
x = c has area 6.
Find the number c > 1 such that the region bounded by the curves y = 1, y = x- 2
,
and x = c has area 1.
Find the number c such that the region bounded by the curves y = x 2 and .r = c has
area 36.
Find the number c > 0 such that the region bounded by the curves y = x 2 and y = ex
has area 9.
Find the value of e between - 1 and 2 such that the area of the region bounded by the
lines y = - x, y = 2x, and y = 1 + ex is a minimum.
Find the value of e such that the line y = e bisects the region bounded by the curves
y = x 2 and y = 1.
Find the value of e such that the line y = ex bisects the region bounded by the x-axis
and the curve y = x - x2
.
4.6 NUMERICAL INTEGRATION
In numerical integration, one computes an approximate value for the definite
integral rather than finding an exact value. In this section we shall present two
methods of numerical integration, called the Trapezoidal Rule and Simpson's Rule.
The Fundamental Theorem of Calculus gives us a method of computing
the definite integral of a given continuous function .f from a to b. The method is to
find, by trial and error, an antiderivative F of .f and then to use the equation f .f(t) dt = F(b) - F(a).
When the method works, it provides an exact value for the integral. However, the
method succeeds only if the antiderivative happens to be a function that can be
described in a simple way. For many integrals one cannot find a formula for the
antiderivative, and the method fails. Such integrals can still be computed approximately
using numerical integration.
The Trapezoidal Rule and Simpson's Rule can always be applied and do
not use the antiderivative. They are easy to carry out on a computer or hand calculator.
We already discussed one method of approximating the definite integral in Section 4.1,
the Riemann sum. The Trapezoidal Rule is a modified form of the Riemann sum,
which gives a much closer approximation for a given amount of effort. Simpson's
Rule is a further modification that gives still better approximations.
4.6 NUMERICAL INTEGRATION 225
Let f be a continuous function on an interval J, and let a < b in J. By
definition, for each positive infinitesimal dx the definite integral
ff(x)dx
is the standard part of the infinite Riemann sum
b L j(x) dx,
a f f(x) dx = s{t f(x) dx].
In Section 4.1, examples were worked out to show that the finite Riemann sums
become very close to the definite integral when .1x is small; that is, the finite Riemann
sums approximate the definite integral. In Section 4.2, we saw that the definite
integral is the limit of the finite Riemann sums as .1x--> o+:
J
b b
a j(x) dx = &!~+ ~ f(x) .1x.
The Riemann sum, which is a sum of areas of rectangles, is a rather inefficient
approximation of the definite integral. We can usually get a much closer approximation
with the same amount of work by adding up areas of trapezoids instead of
rectangles, forming the Trapezoidal Rule suggested by Figure 4.6.1. The Trapezoidal
Rule also provides a formula, called an error estimate, which tells us how close the
approximation is to the exact value of the definite integral.
a b a b
Riemann Sum Trapezoidal Approximation
Figure 4.6.1
Choose a positive integer n and divide the interval [a, b] into n subintervals
of equal length .1x = (b - a)jn. The partition points are a = x 0 , x 1 , x 2 , •.. , x .. = b.
The trapezoidal approximation is the area of the region under the broken line connecting
the points
(x0 ,J(x0 )), (x 1 ,f(x1)), ... , (x .. ,f(x .. )).
Since all of these points lie on the curve y = f(x), the broken line closely follows the
curve. So one would expect the area of the region under the broken line to closely
approximate the area under the curve.
Consider a single subinterval [xm, xm+ 1] of width .1x. The region under the
line segment connecting the two points
226 4 INTEGRATION
is a trapezoid and its area is
f(x,) + f(xm+ 1) L'ix.
2
The sum of the areas of the trapezoids is a modified Riemann sum
I f(x) + j)(x + L'ix) L'ix
a -
= [±f(xo) + f(x1) + f(x2) + • • • + f(x"_1) + !J(x")] L'ix.
We thus make the definition:
DEFINITION
Let L'ix = (b - a)jn evenly divide b -a. Then by the trapezoidal approximation
to the definite integral f~f(x) dx we mean the sum
~, f(x) + f(x + L'ix) 1 1
L
2
L'ix = [zf(xo) + f(x1) + • • • + f(x"_ Il + If(xu}] L'ix.
a
The Trapezoidal Approximation of an integral J~ f(x) dx can be computed
very efficiently on most hand calculators. First compute the sum
!f(xo) + f(x1) + f(x2) + • • • + !f(x")
by cumulative addition. Then multiply this sum by L'ix to obtain the Trapezoidal
Approximation.
THEOREM 1
For a continuous jimction f on [a, b ], the trapezoidal approximation approaches
the definite integral as L'ix--> o+, that is,
fb f(x) dx = lim I f(x) + f(x + L'ix) L'ix.
a ti.x----0 + a 2
PROOF Comparing the formulas for the trapezoidal approximation and the Riemann
sum, we see that
b f(x) + f(x + L'ix) b
L • •
2
L'ix = Lf(x) L'ix + C!f(x.J - tf(xo)) L'ix.
a a
For dx positive infinitesimal, the extra term
(!f(xH) - !f(x0 )) dx
is infinitely small. It follows that
I f(x) + f(x + dx) dx ;::::: I f(x) dx ;::::: fb f(x) dx.
a 2 a a
4.6 NUMERICAL INTEGRATION 227
From a practical standpoint, it is desirable to have a good estimate of error.
We shall first work an example and then state a theorem which gives an error estimate
for the trapezoidal approximation.
EXAMPLE 1 Approximate the definite integral
f jl"+?dx.
Use the trapezoidal approximation with .ilx = i. We first make a table of
values of jl"+?. The graph is drawn in Figure 4.6.2.
Figure 4.6.2
X
x 0 = 0
xi=!
x2 = ~
X3 = i
x4 =!
x 5 = 1
y
)'=~ ./
1.4 /
1.2 _,__../
l.O...L----
J. 1 ~ dx -~ 1.1501
(I
l. 1. 3
5 5
J1"+?
1
Jim
JI!&
J1J§
JIM
I j2
I
! I
~to
four places
1.0000
1.0198
1.0770
1.1662
1.2806
1.4142
X
term in trapezoidal
approximation
0.5000 = tf(xo)
1.0198 = f(x 1 )
1.0770 = f(x 2 )
1.1662 = f(x 3)
1.2806 = f(x4 )
0.7071 = t/(x5)
5.7507 = total
Thus, tf(x0) + f(x 1) + f(x2) + f(x 3) + f(x4) + tf(x5) = 5.7507. Since
.ilx = t, the trapezoidal approximation is
(5.7507). t = 1.1501, f J1"+? dx ~ 1.1501.
The trapezoidal approximation can be made as close to the definite integral
as we want by taking .ilx small. From a practical standpoint, however, it is helpful
to know how small we should take .ilx in order to be sure of a given degree of accuracy.
For instance, suppose we need to know the definite integral to three decimal places.
How small must we take .ilx in our trapezoidal approximation? The answer is given
228 4 INTEGRATION
by the Trapezoidal Rule, which gives an error estimate for the trapezoidal
approximation.
The error in the trapezoidal approximation is the absolute value of the
difference between the trapezoidal sum and the definite integral,
error = It f(x) + f~x + Ll.x) In - f f(x) dx I•
An error estimate for the trapezoidal approximation is a function E(Ll.x), which is
known to be greater than or equal to the error.
Thus if E(Ll.x) is an error estimate, the trapezoidal sum is within E(Ll.x) of
the definite integral. If we want to be sure that the trapezoidal approximation is
accurate to three decimal places-i.e., the error is less than 0.0005-we choose
Ll.x so that E(Ll.x) :::;; 0.0005. We are now ready to state the Trapezoidal Rule.
TRAPEZOIDAL RULE
Let f be a function whose second derivative f" exists and has absolute value
at most M on a closed interval [a, b ],
lf"(x)l:::;; M for a:::;; x:::;; b.
If Ll.x evenly divides b - a, then the trapezoidal approximation of the definite
integral off has the error estimate
b-a
12M(Ll.x)2.
That is,
I
± f(x) + f(x + Ll.x) L'lx - rb f(x) dx I :::;; b - a M(Ll.x)2.
a 2 .Ja 12
The proof is omitted.
EXAMPLE 1 {Concluded) We let f(x) = Jl + x 2
. Then
f
., X
. (x)= v~ • 1 + x2
" jl~i- x2/fl+? 1
J (x) = I + x2 = (I + x2)3i2.
Therefore I f"(x) I :::;; I for all x in [0, I]. We take M = I and use the error
estimate given by the Trapezoidal Rule,
b-a 2 1 (I) 2 I
12M(Ll.x) =12• 1• 5 = 300'
Thus our approximation is within an accuracy of I/300,
I f Jl + x 2 dx - 1.1501 :::;; I/300 ~ 0.0033.
This shows that the integral is, at least, between 1.146 and 1.154.
4.6 NUMERICAL INTEGRATION 229
In this particular example we can even conclude that the integral is between
1.146 and 1.150 (rounded off to three places). That is, the integral is less than its
trapezoidal approximation. This is because the second derivative f"(x) = (1 + x2
)-
3
'
2
is always greater than 0, whence the curve is concave upwards and therefore y = f(x)
is always less than or equal to the broken line used in the trapezoidal approximation.
Actually, the value to three places is 1.148. This can be found by taking ~x = /0 .
EXAMPLE 2 Consider the integral
Let
rl ~dx = n/2.
f(x)=~.
By Theorem 1, we have
.
1 f(x) + f(x + ~x) n
hm I 2 ~x = -2.
Ax--+0+ -1
However, the Trapezoidal Rule fails to give an error estimate in this case
because f'(x) is discontinuous at x = ± 1.
We now turn to Simpson's Rule, for which the number of subintervals n
must be even. As before, we divide the interval [a, b] into n subintervals of equal
length ~x with the n + 1 partition points
a = x0 , x 1, ... , x, =b.
We shall use subintervals of length 2 ~x rather than ~x. On each of the n/2 subintervals
of length 2 ~x we approximate the curve y = f(x) by a parabolic arc that meets the
curve at both endpoints and the midpoint of the subinterval, as shown in Figure 4.6.3.
We then add up the areas under each ofthe parabolic arcs to obtain an approximation
to the area under the curve, which is the definite integral. We begin with a lemma that
gives a formula for the area of the region under one parabolic arc.
Figure 4.6.3
LEMMA
The area of the region under the parabola through three points (u, r), (u + h, s),
and (u + 2h, t) (shown in Figure 4.6.4) is
(u,r)
Figure 4.6.4 l/ l/ + h II + 2h
h
3 (r + 4s + t).
The lemma is proved at the end of this section. Using the lemma, we find
that the area of the region under one parabolic arc from xk to xk + 2 is
It follows that the sum of the n/2 regions under the parabolic arcs is a modified
Riemann sum, t [f(x) + 4f(x + ~x) + f(x + 2 L'.x)J t.x
t.x
= 3 {[f(xo) + 4f(xl) + f(xJ] + [f(x2) + 4f(x 3) + j(x4)] + • • •}
t.x
= 3 [f(x0 ) + 4f(x 1) + 2f(x2 ) + 4f(x 3 ) + 2f(x4) + • • • + 4f(x11 - 1) + f(x")].
This modified Riemann sum is Simpson's approximation to the definite
integral. Note the sequence of coefficients,
1, 4, 2, 4, 2, ... ' 2, 4, 1.
Like the trapezoidal approximation, it is easily computed on a computer or hand
calculator.
THEOREM 2
For a continuous function f on [a, b], Simpson's approximation approaches
the definite integral as t.x --> 0 +,
J
b t.x
f(x) dx = lim - [f(x0 ) + 4f(x 1) + 2f(x2 ) + 4f(x3 ) + • • • + f(x")].
a dx~o• 3
Simpson's approximation is almost as easy to calculate as the trapezoidal
approximation, but is much more accurate. Simpson's Rule is an error estimate
that involves the fourth derivative of the function and the fourth power of t.x.
4.6 NUMERICAL INTEGRATION 231
SIMPSON'S RULE
Suppose the function f has a fourth derivative on the interval [a, b] that has
absolute value at most M,
lf<41(x)l ~ M for a~ x ~b.
If [a, b] is divided into an even number of subintervals of length Ax, then
Simpson's approximation to the definite integral has the error estimate
b-a
18QM(Ax)4
•
EXAMPLE 3 Use Simpson's Rule with Ax= 0.25 to approximate the integral
A = Ll e-x,/2 dx
and find the error estimate.
The curve is the normal (bell-shaped) curve used in statistics, shown in
Figure 4.6.5. We are to divide the interval [0, 1] into four subintervals of
equal length Ax = 0.25. The following table shows the values of x and y
and the coefficient to be used in Simpson's approximation for each partition
point.
y
0 X
Figure 4.6.5 Example 3
X e-xl/2 Coefficient
0.0 1.000000 1
0.25 0.969233 4
0.5 0.882496 2
0.75 0.754840 4
1.0 0.606531
The sum used in the Simpson approximation is then
[1.000000 + 4. (0.969233) + 2. (0.882496) + 4. (0.754840) + 0.606531]
= 10.267816
232 4 INTEGRATION
To get the Simpson approximation, we multiply this sum by b.x/3:
s = (10.267816). (0.25)/3 = 0.855651.
To find the error estimate we need the fourth derivative of
The fourth derivative can be computed as usual and turns out to be
y<4) == (x4 - 6x2 + 3)e- xl/2.
On the interval [0, 1], yl 4
> is decreasing because both x 4
- 6x2 + 3 and
- x 2 /2 are decreasing, and therefore y< 4
> has its maximum value at x = 0
and its minimum value at x == 1,
maximum:
minimum:
yC4l(O) = 3
y(4l(l) = -1.213061
The maximum value of the absolute value I yr 4
> I is thus M = 3. The error
estimate in Simpson's Rule is then
b
1 ~O a (b.x)4 M =
1 ~O • (0.25)4
• 3 = 0.000065.
This shows that the integral is within 0.000065 of the approximation; that is,
fe-x 212 dx = 0.855651 ± 0.000065,
or using inequalities,
0.855586 :-:::; fe-x 212 dx :-:::; 0.855716.
For comparison, a more accurate computation with a smaller b.x shows
that the actual value to six places is
fe-x 212 dx = 0.855624.
The Trapezoidal Rule for this integral and the same value of b.x = 0.25
give an approximate value of 0.85246 for the integral and an error estimate
of 0.00521.
PROOF OF THE LEMMA The algebra is simpler if they-axis is drawn through the
second point, so that u + h = 0, and the three points have coordinates
(-h, r), (0, s), (h, t).
Suppose the parabola has the equation y = ax 2 + bx + c. Then the area
under the parabola is
A= fh(ax2 + bx + c)dx
ax3 bx2 ]h =-+-+ex
3 2 -h
2
= 1ah 3 + 2ch.
4.6 NUMERICAL INTEGRATION 233
When we substitute the coordinates of the three points ( -h, r), (0, s), (h, t)
into the equation for the parabola, we obtain the three equations
r = ah2
- bh + c,
s = c,
t = ah2 + bh + c.
Add the first and third equations and solve for a:
r + t = 2ah2 + 2c
r+t-2c
a = 2h2
Finally, substitute the above expression for a and s for c in the equation
for the area:
2
A= 3ah3 + 2ch
r+t-2c 2 -----:-:~-•-• 113 + 21c1
2h2 3
r + t- 2c + 6c
= •h
3
h
= 3 (r + 4c + t).
h = 3 (r + 4s + t).
PROBLEMS FOR SECTION 4.6
Approximate the integrals in Problems 1-20 using (a) the Trapezoidal Rule and (b) Simpson's
Rule. When possible, find error estimates. If a hand calculator is available, do the problems
again with L'.x = 0.1.
1 S:xdx, L'.x = 0.5 2 Iaz x 3dx, L'.x = 0.5
3 f~dx, L'.x = 0.25 4 r1-d x, L'.x = 0.5
1 X r 5 -1- 2 6 f>Jx+ldx, L'.x = 0.25 dx, L'.x = 0.25
1 1 +X r 7 --Xd x, L'.x = 0.5 8 fP"+Idx, L'.x =±
1 X+ 1
9 fJx4 + 1 dx, L'.x = ± 10 r~dx, L'.x = 0.5
11 r --1 dx
o X+ 1 '
L'.x = 1 12 r2 --1 dx
0 2x+3'
L'.x = 2
13 r3 1
1 x+Jx
dx, L'.x = 3 14 ---dx r 1
0 2+Jx '
L'.x = 1
15
17
19
0 21
0 22
D 23
D 24
D 25
f sin B dB, L'lx = n/2, n/10
{e•' dx, L'lx = ±
f In x dx, L'lx = ±
16
18
20
f sin2 B dB, L'lx = n/2, n/10
{ex' dx, L'lx = ±
f In (1/x) dx, L'lx = ±
Let f be continuous on the interval [a, b] and let L'lx = (b - a)fn where n is a positive
integer. Prove that the trapezoidal sum is equal to the Riemann sum plus ±(j(b) -
f(a)) L'lx, that is,
ttCJ(x) + f(x + L'lx))L'Ix =(tf(x)L'Ix) + ±{_f(b)- f(a)) L'lx.
Show that if f(a) = f(b) then the trapezoidal sum and Riemann sum are equal.
Prove that for a linear function f(x) = kx + c, the trapezoidal sum is exactly equal to
the integral.
Show that if f(x) is concave downward, f"(x) > 0, then the trapezoidal sum is less
than the definite integral ofj(x).
Show that for a quadratic function f(x) = ax 2 + bx + c, Simpson's approximation
is equal to the definite integral.
Show that for a cubic functionf(x) = ax 3 + bx2 +ex+ d, Simpson's approximation
is still equal to the definite integral.
EXTRA PROBLEMS FOR CHAPTER 4
2
3
Evaluate Ii -4 L'lx, L'lx = 1/4
X
Evaluate 'L<.;.'t1 0 21 L'lx,
X
Evaluate I~ 3 2x L'lx,
L'lx = 2
L'lx = 1
4 Evaluate I~ xfi+l L'lx, L'lx = 1/2
5 If F'(x) = 1/(2x - 1)2 for all x =f. 1/2, find F(2)- F(1).
6 If G'(l) = }4t+1 for all t > - 1/4, find G(2) - G(O).
7 A particle moves with velocity v = (3 + 2jt)2
• How far does it move from times t0 = 1
to I 1 = 5?
8 A particle moves with velocity v = t2 ~•How far does it move from times t0 = 1
tot 1 =4?
9 A particle moves with velocity v = (1 + 1)(21 + 3). If it has position y0 = 0 at time
t = 0, find its position at time t = 10.
10 A particle moves with acceleration a = 1/14
. If it has velocity t>0 = 4 and position y0 = 2
at time 1 = 1, find its position at timet = 3.
11 Find the area of the region under the curve y = 1/ fi, 1 -:;; x -:;; 4.
12 Find the area of the region under the curve y = fi - xfi, 0 -:;; x -:;; 1.
In Problems 13-30, evaluate the integral.
13 J (1 - x)(2 + 3x) dx 14
EXTRA PROBLEMS FOR CHAPTER 4 235
15 f (x2: 1)3 dx 16 J (4x + 1) 113 dx
17 f (u/~)du 18 J x- 2)2+x-1 dx
19 f cJ2t+l- J2t=l) ac 20 f 2x + 1
(x + 4)3 dx
21 f yjY+ldy 22 J (1 - p)-4 dx
23 J cos G) dx 24 J Jxsinpdx
25 J e-' dt 26 f -t +- d1 t
t- 1
27 f(y + JY)dy 28 J:(x/~)dx
29 {
1
e4
x dx 30 fx sin (x2) dx
31 Differentiate r jt3+2 dt 32 fx Differentiate
0
(t 2 f(t 2
- l))dt
33 Differentiate r p~ dx 34 Differentiate r2
(1/(x + p)) dx
35 Find the function F such that F'(x) = x - 1 for all x, and the minimum value of F(x) is b .•
36 Find the function F such that F"(x) = x for all x, F(O) = 1, and F(1) = 1.
37 Find the fun.ction F such that F"(x) = 6 for all x, F(x) has a minimum at x = 1, and the
minimum value is 2.
38 Find all functions F such that F"(x) = I + x- 3 for all positive x.
D 39 Find the function F such that
F'(x) = {0
ifx < 0
X ifx 2 0
and F(O) = 1.
D 40 Find the value of b such that the area of the region under the curve y = x(b - x),
0 ::::; x ::::; b, is 1.
D 41 Supposejis increasing for a ::::; x ::::; b, and ~x = (b- a)fn where n is a positive integer.
Show that It f(x) ~x- f f(x) dx I:; [f(b)- f(a)] ~x
D 42 Suppose f is continuous for a ::::; x ::::; b. Show that If f(x) dx I:; f lf(x)l dx.
D 43 Find the area of the top half of the ellipse x 2 fa 2 + y2 fb 2 = 1 using the formula
n = 2f~ 1 jl7du.
D 44 Evaluates~ I (1 - x)312(1 + x) 112 dx using the formula n = 2f~ I J17 du.
D 45 Find dyfdx if y = f~ xf(t) dt.
D 46 Suppose j(t) is continuous for all t and let G(x) = g (x - t)f(t) dt. Prove that
G"(x) = f(x).
D 47 Prove that for any continuous functions f and g,
4 INTEGRATION
2ABj• b f(x)g(x) dx-<; A2 fb j(x)2 dx + B2 lb g(x) 2 dx.
" a a
D 48 Prove Schwartz' Inequality, r f(x)g(x} dx -<; Jfb ~ (x~x -f• b g(x)2 dx.
~~~ u a
Hint: Use the preceding problem.
D 49 Suppose f is continuous and dx is positive infinitesimal. Show that
II> fb f(x + ! dx) dx ~ f(x) dx.
a a
Hint: For each positive real c,
f(x)- c <f(x + ~dx) <f(x) +c.
Use this to show that
Cf(x) dx - c(b -a) < ± f(x + ~ dx) dx < fbJ(x) dx + c(b -a) .
.. a a 2 a
D 50 Suppose j is continuous, 11 is an integer, and dx is positive infinitesimal. Prove that
Ib f(x + fb 11 dx) dx ~ .f(x) dx.
a a
LIMITS,
ANALYTIC GEOMETRY,
AND APPROXIMATIONS
INFINITE LIMITS
Up to this point we have studied three types of limits:
lim f(x) = L