前置知识:变限积分求导
习题1
已知 F ( x ) = ∫ x 2 x sin t 2 d t F(x)=\int_x^{2x}\sin t^2dt F(x)=∫x2xsint2dt,求 F ′ ( x ) F'(x) F′(x)
解:
F ′ ( x ) = sin ( 4 x 2 ) ⋅ 2 − sin ( x 2 ) = 2 sin ( 4 x 2 ) − sin ( x 2 ) \qquad F'(x)=\sin(4x^2)\cdot 2-\sin(x^2)=2\sin(4x^2)-\sin(x^2) F′(x)=sin(4x2)⋅2−sin(x2)=2sin(4x2)−sin(x2)
习题2
已知函数 f ( x ) f(x) f(x)连续, F ( x ) = ∫ 2 x x 2 f ( x ) d x F(x)=\int_{2x}^{x^2}f(x)dx F(x)=∫2xx2f(x)dx,求 F ′ ( x ) F'(x) F′(x)
解:
\qquad 原式 = f ( x 2 ) ⋅ 2 x − f ( 2 x ) ⋅ 2 = 2 x f ( x 2 ) − 2 f ( 2 x ) =f(x^2)\cdot 2x-f(2x)\cdot 2=2xf(x^2)-2f(2x) =f(x2)⋅2x−f(2x)⋅2=2xf(x2)−2f(2x)
习题3
已知 F ( x ) = x 2 ∫ e x ln t d t F(x)=x^2\int_e^x\ln tdt F(x)=x2∫exlntdt,求 f ′ ( e ) f'(e) f′(e)
解:
F ′ ( x ) = 2 x ∫ e x ln t d t + x 2 ln x \qquad F'(x)=2x\int_e^x\ln tdt+x^2\ln x F′(x)=2x∫exlntdt+x2lnx
F ′ ( e ) = 2 x ⋅ 0 + e 2 ⋅ 1 = e 2 \qquad F'(e)=2x\cdot 0+e^2\cdot 1=e^2 F′(e)=2x⋅0+e2⋅1=e2
习题4
计算 lim x → 0 x − ∫ 0 x e t 2 d t x 3 \lim\limits_{x\to 0}\dfrac{x-\int_0^xe^{t^2}dt}{x^3} x→0limx3x−∫0xet2dt
解:
\qquad 由洛必达法则,原式 = lim x → 0 1 − e x 2 3 x 2 = lim x → 0 − 2 x e x 2 6 x = − lim x → 0 e x 2 3 = − 1 3 =\lim\limits_{x\to 0}\dfrac{1-e^{x^2}}{3x^2}=\lim\limits_{x\to 0}\dfrac{-2xe^{x^2}}{6x}=-\lim\limits_{x\to 0}\dfrac{e^{x^2}}{3}=-\dfrac 13 =x→0lim3x21−ex2=x→0lim6x−2xex2=−x→0lim3ex2=−31