变限积分求导

前置知识:牛顿-莱布尼茨公式

介绍

形如

G ( x ) = ∫ v ( x ) u ( x ) f ( t ) d t G(x)=\int_{v(x)}^{u(x)}f(t)dt G(x)=v(x)u(x)f(t)dt

的积分称为变限积分。

牛顿-莱布尼茨公式可得,

G ′ ( x ) = F ′ ( u ( x ) ) u ′ ( x ) − F ′ ( v ( x ) ) v ′ ( x ) = f ( u ( x ) ) u ′ ( x ) − f ( v ( x ) ) v ′ ( x ) G'(x)=F'(u(x))u'(x)-F'(v(x))v'(x)=f(u(x))u'(x)-f(v(x))v'(x) G(x)=F(u(x))u(x)F(v(x))v(x)=f(u(x))u(x)f(v(x))v(x)

那么我们就可以用这个式子来求变限积分的导数了。


例题1

F ( x ) = ∫ 1 x 2 e t d t F(x)=\int_1^{x^2}e^tdt F(x)=1x2etdt,求 F ′ ( x ) F'(x) F(x)

解:
F ′ ( x ) = e x 2 ⋅ 2 x − e 1 ⋅ 0 = 2 x e x 2 \qquad F'(x)=e^{x^2}\cdot 2x-e^1\cdot 0=2xe^{x^2} F(x)=ex22xe10=2xex2


例题2

计算 lim ⁡ x → 0 ∫ ∫ 0 x sin ⁡ t 2 d t x 2 ln ⁡ ( 1 + x ) \lim\limits_{x\to 0}\int\dfrac{\int_0^x\sin t^2dt}{x^2\ln(1+x)} x0limx2ln(1+x)0xsint2dt

解:
\qquad 无穷小代换,原式 = lim ⁡ x → 0 ∫ ∫ 0 x sin ⁡ t 2 d t x 2 ⋅ x =\lim\limits_{x\to 0}\int\dfrac{\int_0^x\sin t^2dt}{x^2\cdot x} =x0limx2x0xsint2dt

\qquad 洛必达法则,原式 = lim ⁡ x → 0 sin ⁡ x 2 3 x 2 = lim ⁡ x → 0 x 2 3 x 2 = 1 3 =\lim\limits_{x\to 0}\dfrac{\sin x^2}{3x^2}=\lim\limits_{x\to 0}\dfrac{x^2}{3x^2}=\dfrac 13 =x0lim3x2sinx2=x0lim3x2x2=31

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转载自blog.csdn.net/tanjunming2020/article/details/131260878
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