前置知识：定积分求曲线的弧长
习题1
已知 f ( x ) = 2 3 x 3 2 f(x)=\dfrac 23x^{\frac 32} f(x)=32x23，求 y = f ( x ) y=f(x) y=f(x)在 [ 0 , 1 ] [0,1] [0,1]上的弧长。
解：
L = ∫ 0 1 1 + f ′ ( x ) 2 d x = ∫ 0 1 1 + x d x \qquad L=\int_0^1\sqrt{1+f'(x)^2}dx=\int_0^1\sqrt{1+x}dx L=∫011+f′(x)2dx=∫011+xdx
= ∫ 0 1 1 + x d ( x + 1 ) = 2 3 ( 1 + x ) 3 2 ∣ 0 1 = 2 3 ⋅ 2 3 2 − 2 3 \qquad\quad =\int_0^1\sqrt{1+x}d(x+1)=\dfrac 23(1+x)^{\frac 32}\bigg\vert_0^1=\dfrac 23\cdot 2^{\frac 32}-\dfrac 23 =∫011+xd(x+1)=32(1+x)23 01=32⋅223−32
习题2
已知 f ( x ) = ln ( 1 − x 2 ) f(x)=\ln(1-x^2) f(x)=ln(1−x2)，求 y = f ( x ) y=f(x) y=f(x)在 [ 0 , 1 2 ] [0,\dfrac 12] [0,21]上的弧长。
解：
L = ∫ 0 1 2 1 + f ′ ( x ) 2 d x = ∫ 0 1 2 1 + ( − 2 x 1 − x 2 ) 2 d x \qquad L=\int_0^{\frac 12}\sqrt{1+f'(x)^2}dx=\int_0^{\frac 12}\sqrt{1+(\dfrac{-2x}{1-x^2})^2}dx L=∫0211+f′(x)2dx=∫0211+(1−x2−2x)2dx
= ∫ 0 1 2 1 + 4 x 2 1 − 2 x 2 + x 4 d x = ∫ 0 1 2 ( 1 + x 2 ) 2 ( 1 − x 2 ) 2 d x \qquad\quad =\int_0^{\frac 12}\sqrt{1+\dfrac{4x^2}{1-2x^2+x^4}}dx=\int_0^{\frac 12}\sqrt{\dfrac{(1+x^2)^2}{(1-x^2)^2}}dx =∫0211+1−2x2+x44x2dx=∫021(1−x2)2(1+x2)2dx
= ∫ 0 1 2 1 + x 2 1 − x 2 d x = − ∫ 0 1 2 1 + x 2 x 2 − 1 d x \qquad\quad =\int_0^{\frac 12}\dfrac{1+x^2}{1-x^2}dx=-\int_0^{\frac 12}\dfrac{1+x^2}{x^2-1}dx =∫0211−x21+x2dx=−∫021x2−11+x2dx
= − ∫ 0 1 2 ( 1 + 2 x 2 − 1 ) d x = − ∫ 0 1 2 ( 1 + 1 x − 1 − 1 x + 1 ) d x \qquad\quad =-\int_0^{\frac 12}(1+\dfrac{2}{x^2-1})dx=-\int_0^{\frac 12}(1+\dfrac{1}{x-1}-\dfrac{1}{x+1})dx =−∫021(1+x2−12)dx=−∫021(1+x−11−x+11)dx
= − ∫ 0 1 2 d x + ∫ 0 1 2 1 1 − x d x + ∫ 0 1 2 1 1 + x d x = − 1 + ln ( 1 − x ) ∣ 0 1 2 + ln ( 1 + x ) ∣ 0 1 2 \qquad\quad =-\int_0^{\frac 12}dx+\int_0^{\frac 12}\dfrac{1}{1-x}dx+\int_0^{\frac 12}\dfrac{1}{1+x}dx=-1+\ln(1-x)\bigg\vert_0^{\frac 12}+\ln(1+x)\bigg\vert_0^{\frac 12} =−∫021dx+∫0211−x1dx+∫0211+x1dx=−1+ln(1−x) 021+ln(1+x) 021
= − 1 + ( ln 1 2 − ln 1 ) + ( ln 3 2 − ln 1 ) = ln 3 − 2 ln 2 − 1 \qquad\quad =-1+(\ln \dfrac 12-\ln 1)+(\ln\dfrac 32-\ln 1)=\ln 3-2\ln 2-1 =−1+(ln21−ln1)+(ln23−ln1)=ln3−2ln2−1
