Definite Integral and Geometry Applications

Definite Integral and Geometry Applications

What is definite integral?

$${\int }_a^{b}f\left(x\right)dx$$ needs to satisfy$a$、$b$ is finite (not infinite) and$f\left(x\right)$ is bounded

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The calculation of the definite integral mainly relies on the NL formula, that is, the Newton-Leibniz formula

Whether the definite integral exists is irrelevant whether the original function exists or not

Skills and test points for definite integral calculation:

1. symmetry
2. geometric meaning
3. Integrand additivity

calculation skills

Can combine graphics, such as ${\int }_0^1\sqrt{1-x^2}$Actually $\frac{1}{4}$unit circle

Using the symmetric interval,Double for even function, 0 for odd functionproperties, which means that it is possible to separate the integrand into even and odd functions, and it is also possible to dismantle the integral interval into several intervals including the symmetric interval

High-power sine-cosine trigonometric functions look at the upper and lower limits of the integral, find a way to use itignition formula, if it is not within the upper and lower limits of the ignition formula, find a way to change it

definite point swap

To change the element, three things need to be changed: the upper and lower limits, the integrand function, and the integral variable

If let $t=g\left(x\right)$ , then the original integral will become$g(b)$，$g\left(a\right)$ , the integrand is$f\left(t\right)$ about$The\; definite\; integral\; of\; t$ , such as:

$${\int }_0^1{x}^2\sqrt{1-x^2}dx\stackrel{x=sint}{\to }{\int }_0^{\frac{}{2}}{\sin }^{2}t{\cos }^{2}tdt$$

The upper and lower limits are $x=sint$ is substituted into$The\; value\; of\; x$ , the integrand function is also a direct replacement, then$dx=d(\sin t)=\cos tdt$

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The processing methods of changing elements and indefinite integrals here are somewhat similar.

interval reproduction formula

$${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$$

To prove it, let $x=a+b-t$ , find out the formula after the substitution, and finally find thatWhat letter is used for the integral value is irrelevant(That is, the formula after the substitution is the same as the previous formula but the letters are different)

[Note]: The use of the interval reproduction formula generally relies on swapping elements, and the typical idea is as follows:

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I took a picture from Zhihu, see the watermark for the source of the picture:

Wallace Formula / Ignition Formula

$${\int }_0^{\frac{}{2}}{\sin }^{n}xdx={\int }_0^{\frac{}{2}}{\cos }^{n}xdx=ignition\; formula$$

$${\int }_0^{}{\sin }^{n}xdx=2{\int }_0^{\frac{}{2}}{\sin }^{n}xdx\to Then\; the\; ignition\; formula$$

$${\int }_0^{}{\cos }^{n}xdx=\{\begin{array}{rlr}0& & nis\; a\; positive\; odd\; number\\ 2{\int }_0^{\frac{}{2}}{\cos }^{n}xd& & nis\; a\; positive\end{array}$$

Definite Integrals of Parametric Equations

See this article: Knowing - Zero Basics High Numbers | Using definite integrals to calculate the plane graphic area of ​​​​parametric equations

If it is in the Cartesian coordinate system, then there is $y=f\left(x\right)$ , the formula for calculating the area at this time is

$$S={\int }_0^{2\pi a}f(x)dx$$

Then to find $y=$The representation of the parametric equation of$f$$($$x$$)\; ,\; namely$Exchange method

To change currencythree changes

1. Change the integral variable: $dx\to dx(t)$
2. Change the upper and lower limits of the integral, according to the t = f ( x ) t=f(x) set when changing the element$t=f(x)$
3. $f\left(x\right)$ to$y(t)$

From this we get:

$$S={\int }_0^{2}y(t)dx(t)={\int }_0^{2}y(t)x^{\prime }(t)dt$$

The reason for the third step lies in the following formula (it can be seen by substituting a specific parameter equation) $$y=f(x)=f[x(t)]=y(t)$$

Back to the original question, bring in the specific value

$$S={\int }_0^{2}a(1-\cos t)a(1-\cos t)dt=a^2{\int }_0^2(1-\cos t{)}^{2}dt$$

$$a^2{\int }_0^2(1-\cos t{)}^{2}dt=a^2{\int }_0^2(1-2\cos t+{\cos }^{2}t)dt$$

Split here, for ${\cos }^{2}t$ using the ignition formula (Wales formula)

Definite Integrals in Polar Coordinates

(Question source Zhang Yu 300 Question 9.2)

Just remember the following formula:

$$\{\begin{array}{rl}x& =r\cos i\\ y& =r\sin \end{array}$$

所以$(x^2+y^2{)}^2=x^2-y^2$ can be reduced to$r^2=\cos 2\theta$ , on the other hand, the polar curve$r=r\left(\theta \right)$ Synchronous coefficient$$S=\frac{1}{2}{\int }_a^{}{r}^2\left(i\right)di$$

The key thing to note is that there is a $\frac{1}{2}$

variable limit integral

A variable limit integral must be continuous as long as it exists

Precise definition of definite integral (sequence limit to definite integral)

I have to mention herePrecise Definition of Definite Integral, because it connects the limit to the integral

$${\int }_a^{b}f(x)dx=\underset{n\to \infty }{\lim }\sum _{i=1}^{n}f(a+\frac{b-a}{n}i)\frac{b-a}{n}$$

In particular, if $\frac{1}{n}$Actually $b=1$、$a=$bat$0\; -\; an\; \backslash frac\{ba\}n$$\frac{b-a}{n}$, then the above formula can be reduced to:

$${\int }_0^{1}f(x)dx=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{i=1}^{n}f(\frac{i}{n})$$

i.e. the last $\frac{1}{n}$用$dx$ replace,$\frac{1}{n}$use $x$ replace

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【Example of Yang Chao】

[Zhang Yu 300 question 8.6]

$$\underset{n\to \infty }{\lim }(\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+...+\frac{1}{\sqrt{n^2+n^2}})=?$$

The denominator of each element can be proposed a $\frac{1}{n}$

$$\frac{1}{\sqrt{n^2+kn}}\to \frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n}}}$$

Then put $\frac{1}{n}$Proposed, at this time the original formula into the following formula ( $b=1$、$a=0$ is obvious):

$$\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{k=1}^n\frac{1}{\sqrt{1+\frac{k}{n}}}={\int }_0^1\frac{1}{\sqrt{1+x}}dx$$

Finally, using the substitution method for the denominator can obtain the result $2\sqrt{2}-2$

to $t$ integral but the integrand contains$x$

例如$$\frac{d}{dx}[{\int }_0^{x}tf(x^2-t^2)dt]$$

令$u=x^2-t^2$ , move the derivative variable out of the integrand through variable substitution, then$-2tdt=ofu$

Then change the upper and lower limits (in fact, bring the original upper and lower limits to $u=x^2-t^2$ ), then the original formula is equal to

$$\frac{1}{2}\frac{d}{dx}[{\int }_0^{x^2}f(u\left)du\right]\stackrel{Variable\; limit\; integral\; derivation\; formula}{\to }xf(x^2)$$

abnormal integral

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