《高等数学》习题3.4选做
Elegia
2021-06-20 23:29:27
求下列各定积分：
1. $\displaystyle \int _{-1}^{1}\frac{x\mathrm{d} x}{\sqrt{5-4x}}$
解：令 $\displaystyle t=\sqrt{5-4x}$，有
$$
\begin{aligned}
& =\int _{3}^{1}\frac{\frac{5-t^{2}}{4}}{t}\mathrm{d}\left(\frac{5-t^{2}}{4}\right)\\
& =\int _{1}^{3}\frac{5-t^{2}}{4t} \cdotp \frac{t}{2}\mathrm{d} t\\
& =\int _{1}^{3}\frac{5-t^{2}}{8}\mathrm{d} t\\
& =\left. \frac{5t-\frac{1}{3} t^{3}}{8}\right| _{1}^{3}\\
& =\frac{1}{6}
\end{aligned}
$$
2. $\displaystyle \int _{0}^{\ln 2} x\mathrm{e}^{-x}\mathrm{d} x$
解：$\displaystyle =-\int _{0}^{\ln 2} x\mathrm{d}\left(\mathrm{e}^{-x}\right) =-\left. x\mathrm{e}^{-x}\right| _{0}^{\ln 2} +\int _{0}^{\ln 2}\mathrm{e}^{-x}\mathrm{d} x=-\frac{\ln 2}{2} -\left. \mathrm{e}^{-x}\right| _{0}^{\ln 2} =\frac{1-\ln 2}{2}$
3. $\displaystyle \int _{0}^{1} x^{2}\sqrt{1-x^{2}}\mathrm{d} x$
解：令 $\displaystyle x=\sin t$，有
$$
\begin{aligned}
& =\int _{0}^{\pi /2}\sin^{2} t\cos t\mathrm{d}(\sin t)\\
& =\frac{1}{4}\int _{0}^{\pi /2}\sin^{2} 2t\mathrm{d} t\\
& =\frac{1}{8}\int _{0}^{\pi /2}( 1-\cos 4t)\mathrm{d} t\\
& =\frac{\pi }{16}
\end{aligned}
$$
5. $\displaystyle \int _{0}^{4}\sqrt{x^{2} +9}\mathrm{d} x$
解：
$$
\begin{aligned}
I & =\left. x\sqrt{x^{2} +9}\right| _{0}^{4} -\int _{0}^{4}\frac{x^{2}}{\sqrt{x^{2} +9}}\mathrm{d} x\\
& =20-\int _{0}^{4}\frac{x^{2} +9-9}{\sqrt{x^{2} +9}}\mathrm{d} x\\
2I & =20+9\int _{0}^{4}\frac{\mathrm{d} x}{\sqrt{x^{2} +9}}\\
I & =10+\frac{9}{2}\left. \left(\ln\left| x+\sqrt{x^{2} +9}\right| \right)\right| _{0}^{4}\\
& =10+\frac{9}{2}\ln 3
\end{aligned}
$$
6. $\displaystyle \int _{0}^{1/2}\frac{x^{2}}{\sqrt{1-x^{2}}}\mathrm{d} x$
解：令 $\displaystyle x=\sin t$，有
$$
\begin{aligned}
& =\int _{0}^{\pi /6}\frac{\sin^{2} t}{\cos t}\mathrm{d}(\sin t)\\
& =\int _{0}^{\pi /6}\frac{1-\cos 2t}{2}\mathrm{d} t\\
& =\frac{\pi }{12} -\left. \frac{\sin 2t}{4}\right| _{0}^{\pi /6}\\
& =\frac{\pi }{12} -\frac{\sqrt{3}}{8}
\end{aligned}
$$
8. $\displaystyle \int _{0}^{3} x\sqrt[3]{1-x^{2}}\mathrm{d} x$
解：令 $\displaystyle t=\sqrt[3]{1-x^{2}}$，也就有 $\displaystyle x=\sqrt{1-t^{3}}$。
$$
\begin{aligned}
& =\int _{1}^{-2} t\sqrt{1-t^{3}}\mathrm{d}\left(\sqrt{1-t^{3}}\right)\\
& =\int _{1}^{-2} t\cdotp \frac{-3}{2} t^{2}\mathrm{d} t\\
& =\left. -\frac{3}{8} t^{4}\right| _{1}^{-2}\\
& =-\frac{45}{8}
\end{aligned}
$$
9. $\displaystyle \int _{-\pi /2}^{\pi /2}\sqrt{\cos x-\cos^{3} x}\mathrm{d} x$
解：先根据偶函数变成一半，然后令 $\displaystyle t=\cos x$，
$$
\begin{aligned}
& =2\int _{1}^{0}\sqrt{t\left( 1-t^{2}\right)}\mathrm{d}(\arccos t)\\
& =2\int _{0}^{1}\sqrt{t}\mathrm{d} t\\
& =\frac{4}{3}
\end{aligned}
$$
10. $\displaystyle \int _{0}^{\pi /2}\cos^{n} 2x\mathrm{d} x$，其中 $\displaystyle n$ 为正整数。
解：对于 $\displaystyle n\geq 2$，有
$$
\begin{aligned}
I_{n} & =\frac{1}{2}\int _{0}^{\pi /2}\cos^{n-1} 2x\mathrm{d}(\sin 2x)\\
& =\left. \frac{1}{2}\cos^{n-1} 2x\sin 2x\right| _{0}^{\pi /2} -\frac{1}{2}\int _{0}^{\pi /2}\sin 2x\mathrm{d}\left(\cos^{n-1} 2x\right)\\
& =\frac{n-1}{2}\int _{0}^{\pi /2}\cos^{n-2} 2x\left( 1-\cos^{2} 2x\right)\mathrm{d} x\\
& =\frac{n-1}{2}( I_{n-2} -I_{n})\\
\frac{n+1}{2} I_{n} & =\frac{n-1}{2} I_{n-2}\\
I_{n} & =\frac{n-1}{n+1} I_{n-2}\\
I_{2k} & =\frac{2k-1}{2k+1} \cdots \frac{1}{3} I_{0}\\
& =\frac{( 2k-1) !!}{( 2k+1) !!} \cdotp \frac{\pi }{2}\\
I_{2k+1} & =\frac{2k}{2k+2} \cdots \frac{2}{4} I_{1} =0
\end{aligned}
$$
11. $\displaystyle \int _{0}^{a}\left( a^{2} -x^{2}\right)^{n/2}\mathrm{d} x$
解：对于 $\displaystyle n >0$，
$$
\begin{aligned}
I_{n} & =\left. x\left( a^{2} -x^{2}\right)^{n/2}\right| _{0}^{a} -\int _{0}^{a} x\mathrm{d}\left[\left( a^{2} -x^{2}\right)^{n/2}\right]\\
& =+n\int _{0}^{a}\left( a^{2} -x^{2}\right)^{( n-2) /2}\left( x^{2} -a^{2} +a^{2}\right)\mathrm{d} x\\
& =na^{2} I_{n-2} -nI_{n}\\
I_{n} & =\frac{n}{n+1} a^{2} I_{n-2}\\
I_{2k} & =\frac{2ka^{2}}{2k+1} \cdots \frac{2a^{2}}{3} I_{0}\\
& =\frac{( 2k) !!}{( 2k+1) !!} a^{2k+1}\\
I_{2k+1} & =\frac{( 2k+1) a}{2k+2} \cdots \frac{3a}{4} I_{1}\\
& =\frac{( 2k+1) !!}{( 2k+2) !!} \cdot \frac{\pi a^{2k+2}}{2}
\end{aligned}
$$
15. $\displaystyle \int _{0}^{\pi /4}\tan^{4} x\mathrm{d} x$
解：令 $\displaystyle t=\tan x$，有
$$
\begin{aligned}
& =\int _{0}^{1} t^{4}\mathrm{d}(\arctan t)\\
& =\int _{0}^{1}\frac{t^{4}}{1+t^{2}}\mathrm{d} t\\
& =\int _{0}^{1}\left( t^{2} -1+\frac{1}{1+t^{2}}\right)\mathrm{d} t\\
& =\frac{\pi }{4} -\frac{2}{3}
\end{aligned}
$$
18. 设 $\displaystyle f( x)$ 在 $\displaystyle [ a,b]$ 上连续，证明 $\displaystyle \int _{a}^{b} f( x)\mathrm{d} x=( b-a)\int _{0}^{1} f[ a+( b-a) x]\mathrm{d} x$
解：做换元 $\displaystyle x=a+( b-a) t$，也就有 $\displaystyle t=\frac{x-a}{b-a}$，有 $\displaystyle \mathrm{l.h.s.} =\int _{0}^{1} f[ a+( b-a) t]\mathrm{d}[ a+( b-a) t] =( b-a)\int _{0}^{1} f[ a+( b-a) t]\mathrm{d} t$，而积分的值和字母无关。
19. 证明：$\displaystyle \int _{0}^{a} x^{3} f\left( x^{2}\right)\mathrm{d} x=\frac{1}{2}\int _{0}^{a^{2}} xf( x)\mathrm{d} x$
解：$\displaystyle \int _{0}^{a} x^{3} f\left( x^{2}\right)\mathrm{d} x=\frac{1}{2}\int _{0}^{a} x^{2} f\left( x^{2}\right)\mathrm{d}\left( x^{2}\right)$，于是换元为 $\displaystyle t=x^{2}$，就有 $\displaystyle =\frac{1}{2}\int _{0}^{a^{2}} tf( t)\mathrm{d} t$。
20. 证明 $\displaystyle \int _{0}^{1} x^{m}( 1-x)^{n}\mathrm{d} x=\int _{0}^{1} x^{n}( 1-x)^{m}\mathrm{d} x$
解：令 $\displaystyle t=1-x$，则有 $\displaystyle \mathrm{l.h.s.} =-\int _{1}^{0}( 1-t)^{m} t^{n}\mathrm{d} t=\int _{0}^{1} t^{n}( 1-t)^{m}\mathrm{d} t$。
21. 利用分部积分公式证明若 $\displaystyle f( x)$ 连续，则 $\displaystyle \int _{0}^{x}\left[\int _{0}^{t} f( x)\mathrm{d} x\right]\mathrm{d} t=\int _{0}^{x} f( t) \cdot ( x-t)\mathrm{d} t$
解：
$$
\begin{aligned}
\mathrm{l.h.s.} & =\left. x\int _{0}^{t} f( x)\mathrm{d} x\right| _{0}^{x} -\int _{0}^{x} t\mathrm{d}\left[\int _{0}^{t} f( x)\mathrm{d} x\right]\\
& =x\int _{0}^{x} f( t)\mathrm{d} t-\int _{0}^{x} tf( t)\mathrm{d} t\\
& =\int _{0}^{x} f( t) \cdot ( x-t)\mathrm{d} t
\end{aligned}
$$
22. 利用换元积分法证明：$\displaystyle \int _{0}^{\pi } xf(\sin x)\mathrm{d} x=\pi \int _{0}^{\pi /2} f(\sin x)\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int _{0}^{\pi /2} xf(\sin x)\mathrm{d} x+\int _{\pi /2}^{\pi } xf(\sin x)\mathrm{d} x\\
& =\int _{0}^{\pi /2} xf(\sin x)\mathrm{d} x+\int _{\pi /2}^{0}( \pi -x) f(\sin( \pi -x))\mathrm{d}( \pi -x)\\
& =\int _{0}^{\pi /2} xf(\sin x)\mathrm{d} x+\int _{0}^{\pi /2}( \pi -x) f(\sin x)\mathrm{d} x\\
& =\pi \int _{0}^{\pi /2} f(\sin x)\mathrm{d} x
\end{aligned}
$$
23. 利用上题结果，求 $\displaystyle \int _{0}^{\pi }\frac{x\thinspace \sin x}{1+\cos^{2} x}\mathrm{d} x$
解：
$$
\begin{aligned}
& =\int _{0}^{\pi }\frac{x\thinspace \sin x}{2-\sin^{2} x}\mathrm{d} x\\
& =\pi \int _{0}^{\pi /2}\frac{\sin x}{2-\sin^{2}}\mathrm{d} x\\
& =-\pi \int _{0}^{\pi /2}\frac{\mathrm{d}(\cos x)}{1+\cos^{2} x}\\
& =\pi \int _{0}^{1}\frac{\mathrm{d} t}{1+t^{2}}\\
& = \pi \arctan t| _{0}^{1}\\
& =\frac{\pi ^{2}}{4}
\end{aligned}
$$
24. 设函数 $\displaystyle f( x)$ 在 $\displaystyle ( -\infty ,+\infty )$ 上连续，以 $\displaystyle T$ 为周期，证明：
(1) 函数 $\displaystyle F( x) =\frac{x}{T}\int _{0}^{T} f( t)\mathrm{d} t-\int _{0}^{x} f( t)\mathrm{d} t$ 也以 $\displaystyle T$ 为周期。
解：
$$
\begin{aligned}
F( x+T) -F( x) & =\frac{x+T-x}{T}\int _{0}^{T} f( t)\mathrm{d} t-\int _{x}^{x+T} f( t)\mathrm{d} t\\
& =\int _{0}^{T} f( t)\mathrm{d} t-\int _{x}^{x+T} f( t)\mathrm{d} t\\
& =0
\end{aligned}
$$
得证。
(2) $\displaystyle \lim _{x\rightarrow +\infty }\frac{1}{x}\int _{0}^{x} f( t)\mathrm{d} t=\frac{1}{T}\int _{0}^{T} f( t)\mathrm{d} t$
解：考虑第一问中的 $\displaystyle F( x)$，由于其是连续函数，在 $\displaystyle [ 0,T]$ 上有界，又是周期函数所以整体有界，有 $\displaystyle | F( x)| < M$。而 $\displaystyle x\cdot \left| \frac{1}{x}\int _{0}^{x} f( t)\mathrm{d} t-\frac{1}{T}\int _{0}^{T} f( t)\mathrm{d} t\right| =| F( x)| < M$。因此对于任意 $\displaystyle \varepsilon >0$，当 $\displaystyle x >\frac{M}{\varepsilon }$ 时即有 $\displaystyle \left| \frac{1}{x}\int _{0}^{x} f( t)\mathrm{d} t-\frac{1}{T}\int _{0}^{T} f( t)\mathrm{d} t\right| < \varepsilon $。
25. 设 $\displaystyle f( x)$ 是以 $\displaystyle T$ 为周期的连续函数，$\displaystyle f( x_{0}) \neq 0$，且 $\displaystyle \int _{0}^{T} f( x)\mathrm{d} x=0$，证明 $\displaystyle f( x)$ 在 $\displaystyle ( x_{0} ,x_{0} +T)$ 内至少有两个根。
解：不妨设 $\displaystyle f( x_{0}) >0$，那么由于 $\displaystyle \int _{x_{0}}^{x_{0} +T} f( x)\mathrm{d} x=\int _{0}^{T} f( x)\mathrm{d} x=0$，接下来考虑是否存在一个点 $\displaystyle x_{1}$ 使得 $\displaystyle f( x_{1}) < 0$。如果不存在，说明 $\displaystyle f( x) =0$（否则积分 $\displaystyle >0$），取两个三等分点即可。如果存在，那么使用中值定理可知存在 $\displaystyle x_{2} \in ( x_{0} ,x_{1}) ,x_{3} \in ( x_{1} ,x_{0} +T)$ 使得 $\displaystyle f( x_{2}) =f( x_{3}) =0$，也就找到了两个根。
26. 求定积分 $\displaystyle \int _{0}^{2m\pi }\frac{\mathrm{d} x}{\sin^{4} x+\cos^{4} x}$，其中 $\displaystyle m$ 为正整数。
解：注意到函数有周期 $\displaystyle \pi $，因此
$$
\begin{aligned}
& =2m\int _{0}^{\pi }\frac{\mathrm{d} x}{\left( 1-\cos^{2} x\right)^{2} +\cos^{4} x}\\
& =2m\int _{0}^{\pi }\frac{\mathrm{d} x}{\left(\sin^{2} x+\cos^{2} x\right)^{2} -2\sin^{2} x\cos^{2} x}\\
& =2m\int _{0}^{\pi }\frac{\mathrm{d} x}{1-\frac{1}{2}\sin^{2} 2x}\\
& =4m\int _{0}^{\pi }\frac{\mathrm{d} x}{2-\frac{1-\cos 4x}{2}}\\
& =2m\int _{0}^{\pi }\frac{\mathrm{d}( 4x)}{3+\cos 4x}\\
& =2m\int _{0}^{4\pi }\frac{\mathrm{d} x}{3+\cos x}\\
& =8m\int _{0}^{\pi }\frac{\mathrm{d} x}{3+\cos x}\\
I & =\int _{0}^{\pi }\frac{\mathrm{d} x}{3+\cos x}\\
& =\int _{0}^{\pi }\frac{\mathrm{d} x}{3+\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\\
& =\int _{0}^{+\infty }\frac{\left( 1+t^{2}\right)\mathrm{d}( 2\arctan t)}{4+2t^{2}}\\
& =\int _{0}^{+\infty }\frac{\mathrm{d} t}{2+t^{2}}\\
& =\left. \frac{1}{\sqrt{2}}\arctan\frac{t}{\sqrt{2}}\right| _{0}^{+\infty }\\
& =\frac{\pi }{2\sqrt{2}}\\
8mI & =2\sqrt{2} m\pi
\end{aligned}
$$