MIT_单变量微积分_19

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1.定积分在对数和几何上的应用

FTC2:
d d x a x f ( t ) d ( t ) = f ( x ) y = 1 x ; L ( x ) = 1 x d t t L ( x ) = 1 x ; L ( x ) = 1 1 d t t = 0 L = 1 x 2 L ( 1 ) = 0 ; L ( 1 ) = 1 \frac{d}{dx}\int_a^xf(t)d(t)=f(x)\\ y'=\frac{1}{x};L(x)=\int_1^x\frac{dt}{t}\\ L'(x)=\frac{1}{x};L(x) = \int _1^1\frac{dt}{t}=0\\ L'' = -\frac{1}{x^2}\\ L(1) = 0; L'(1)=1
L ( e ) = 1 L(e) = 1
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为什么 L ( x ) < 0 , x < 1 L(x) < 0, x < 1 ???

  • L(1) = 0 , L递增
  • L ( x ) = 1 x d t t = x 1 d t t L(x) = \int_1^x\frac{dt}{t}=-\int_x^1\frac{dt}{t}

L ( a b ) = L ( a ) + L ( b ) 1 a b d t t = 1 a d t t + a a b d t t a a b d t t = 1 b a d u a u = 1 b d u u = L ( b ) t = a u , d t = a d u L(ab)=L(a)+L(b)\\ \int_1^{ab}\frac{dt}{t}=\int_1^{a}\frac{dt}{t}+\int_a^{ab}\frac{dt}{t}\\ 由于 \int_a^{ab}\frac{dt}{t}=\int_1^b\frac{adu}{au}=\int_1^b\frac{du}{u}=L(b)\\ t = au, dt = adu

Ex: 0 x e t 2 d t \int_0^xe^{-t^2}dt
F ( x ) = e x 2 , F ( 0 ) = 0 , F ( 0 ) = e 0 2 = 1 F = 2 x e x 2 { < 0 , x > 0 > 0 , x < 0 F'(x)=e^{-x^2},F(0)=0, F'(0)=e^{0^2}=1\\ F''=-2xe^{-x^2}\left\{\begin{matrix} <0&,x > 0 \\ >0&,x<0 \end{matrix}\right.
如图所示: F F'' 图像,虚线为 y = π 2 y=\frac{\sqrt{\pi}}{2} ,为奇函数, F ( x ) = F ( x ) F(-x)=-F(x)

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如图所示: F = e x 2 F'=e^{-x^2}

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lim x F ( x ) = π 2 \lim_{x \to \infty} F(x)=\frac{\sqrt{\pi}}{2}

e r f ( x ) = 2 π 0 x e t 2 d t = 2 z erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt=\frac{2}{z}


扩展的特殊函数:

  • L i ( x ) = 2 x d t l n t p r i m e s ( ) < x Li(x)=\int_2^x\frac{dt}{ln t} \approx primes(素数个数) < x
  • C ( x ) = 0 x c o s ( t 2 ) d t S ( x ) = 0 x s i n ( t 2 ) d t } H ( x ) = 0 x s i n t t d t ( ) \left.\begin{matrix} C(x)=\int_0^xcos(t^2)dt& \\ S(x)=\int_0^xsin(t^2)dt& \end{matrix}\right\}菲涅耳积分\\ H(x)=\int_0^x\frac{sint}{t}dt(傅里叶级数)

Ex:关于曲线积分围城的面积
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A r e a = a b ( f ( x ) g ( x ) ) d x Area=\int_a^b(f(x)-g(x))dx

Ex x = y 2 x=y^2 y = x 2 y=x-2 围成的面积。
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  • (垂直切分)
    A r e a = 0 1 ( x ( x ) ) d x + 1 4 ( x ( x 2 ) ) d x = . . . Area=\int_0^1(\sqrt{x}-(-\sqrt{x}))dx+\int_1^4(\sqrt{x}-(x-2))dx=...
  • (横向切分)

A r e a = 1 2 ( ( y + 2 ) y 2 ) d y = ( y 2 2 + 2 y y 3 3 ) 1 2 = 9 2 Area=\int_{-1}^2((y+2)-y^2)dy=(\frac{y^2}{2}+2y-\frac{y^3}{3})|_{-1}^2=\frac{9}{2}

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