中册 4.2 定积分计算 第32题

解题过程： （1）记 $\int_{1}^{\mathrm{e}} f(x) \mathrm{d} x=a$ ，则 $f(x)=\ln x-a$ ，从而 $$ a=\int_{1}^{\mathrm{e}} f(x) \mathrm{d} x=\int_{1}^{\mathrm{e}}(\ln x-a) \mathrm{d} x=\int_{1}^{\mathrm{e}} \ln x \mathrm{~d} x-a(\mathrm{e}-1)=1-a(\mathrm{e}-1) . $$ 由此得 $\displaystyle a=\frac{1}{\mathrm{e}}$ ．所以 $\displaystyle \int_{1}^{\mathrm{e}} f(x) \mathrm{d} x=\frac{1}{\mathrm{e}}$ ． （2）两边从 0 至 1 积分得 $\int_{0}^{1} f(x) \mathrm{d} x=\int_{0}^{1} 3 x^{2} \mathrm{~d} x+2 \int_{0}^{1} f(x) \mathrm{d} x$ ，所以 $\int_{0}^{1} f(x) \mathrm{d} x=-1$ ． 两边从1至2积分得 $\int_{1}^{2} f(x) \mathrm{d} x=\int_{1}^{2} 3 x^{2} \mathrm{~d} x+2 \int_{0}^{1} f(x) \mathrm{d} x=7-2=5$ ． （3）令 $u=2 x-t$ ，则 $\int_{0}^{1} t f(2 x-t) \mathrm{d} t=-\int_{2 x}^{2 x-1}(2 x-u) f(u) \mathrm{d} u$ 。于是 $$ 2 x \int_{2 x-1}^{2 x} f(u) \mathrm{d} u-\int_{2 x-1}^{2 x} u f(u) \mathrm{d} u=\frac{1}{2} \arctan \left(x^{2}\right) $$ 两边求导得 $\displaystyle 2 \int_{2 x-1}^{2 x} f(u) \mathrm{d} u+4 x(f(2 x)-f(2 x-1))-2(2 x f(2 x)-(2 x-1) f(2 x-1))=\frac{x}{1+x^{4}}$ ． 将 $x=1, f(1)=1$ 代人上式得 $\displaystyle \int_{1}^{2} f(x) \mathrm{d} x=\frac{5}{4}$ ． （4）令 $u=x-t$ ，则 $$ \int_{0}^{x} t f(x-t) \mathrm{d} t=\int_{x}^{0}(x-u) f(u)(-\mathrm{d} u)=\int_{0}^{x}(x-u) f(u) \mathrm{d} u=x \int_{0}^{x} f(u) \mathrm{d} u-\int_{0}^{x} u f(u) \mathrm{d} u=1-\cos x $$ 两边对 $x$ 求导得 $$ \int_{0}^{x} f(u) \mathrm{d} u+x f(x)-x f(x)=\sin x $$ 故 $$ \int_{0}^{x} f(u) \mathrm{d} u=\sin x $$ 令 $\displaystyle x=\frac{\pi}{2}$ ，则 $\displaystyle \int_{0}^{\frac{\pi}{2}} f(u) \mathrm{d} u=\int_{0}^{\frac{\pi}{2}} f(x) \mathrm{d} x=1$ ． （5）由 $F(x) f(x)=\cos 2 x$ 得 $\mathrm{d} F^{2}(x)=2 F(x) f(x) \mathrm{d} x=2 \cos 2 x \mathrm{~d} x$ 。则 $$ F^{2}(x)=\sin 2 x+C $$ 由 $F(0)=1$ 得 $C=1$ ，于是 $F^{2}(x)=\sin 2 x+1$ ．即 $F(x)=\sqrt{\sin 2 x+1}$ ． 记 $\displaystyle x_{0}=\frac{3 \pi}{4}$ ．当 $\sin 2 x+1 \neq 0$ ，即 $x \neq x_{0}$ 时， 因 $$ \begin{aligned} f(x)=F^{\prime}(x) & =\frac{\cos 2 x}{\sqrt{\sin 2 x+1}}=\frac{\cos ^{2} x-\sin ^{2} x}{|\sin x+\cos x|}=\frac{\sin x+\cos x}{|\sin x+\cos x|}(\cos x-\sin x) \\ & =-\sqrt{2} \frac{\sin x+\cos x}{|\sin x+\cos x|} \sin \left(x-\frac{\pi}{4}\right) \end{aligned} $$ $$ \begin{aligned} & \lim _{x \rightarrow x_{0}^{+}} f(x)=-\sqrt{2} \lim _{x \rightarrow x_{0}^{+}} \frac{\sin x+\cos x}{|\sin x+\cos x|} \sin \left(x-\frac{\pi}{4}\right)=\sqrt{2} \\ & \lim _{x \rightarrow x_{0}^{-}} f(x)=-\sqrt{2} \lim _{x \rightarrow x_{0}^{-}} \frac{\sin x+\cos x}{|\sin x+\cos x|} \sin \left(x-\frac{\pi}{4}\right)=-\sqrt{2} \end{aligned} $$ 所以 $x=x_{0}$ 为 $f(x)$ 的跳跃间断点，$f(x)$ 在 $[0, \pi]$ 可积．故 $$ \begin{aligned} \int_{0}^{\pi}|f(x)| \mathrm{d} x & =\int_{0}^{\pi}\left|F^{\prime}(x)\right| \mathrm{d} x=\sqrt{2} \int_{0}^{\pi}\left|\sin \left(x-\frac{\pi}{4}\right)\right| \mathrm{d} x=\sqrt{2} \int_{-\frac{\pi}{4}}^{\frac{3 \pi}{4}}|\sin t| \mathrm{d} t \\ & =\sqrt{2}\left(-\int_{-\frac{\pi}{4}}^{0} \sin t \mathrm{~d} t+\int_{0}^{\frac{3 \pi}{4}} \sin t \mathrm{~d} t\right)=2 \sqrt{2} \end{aligned} $$ （6）$\displaystyle f^{\prime}(\tan x)=\sin x=\frac{\tan x}{\sqrt{1+\tan ^{2} x}}$ ．令 $t=\tan x$ ，则 $\displaystyle f^{\prime}(t)=\frac{t}{\sqrt{1+t^{2}}}$ ，故 $$ f(t)=\int \frac{t}{\sqrt{1+\mathrm{t}^{2}}} \mathrm{~d} t=\sqrt{1+t^{2}}+C $$ 由 $f(0)=1$ 得 $C=0$ ，故 $f(t)=\sqrt{1+t^{2}}$ ．于是 $$ \int_{0}^{1} x f(x) \mathrm{d} x=\int_{0}^{1} x \sqrt{1+x^{2}} \mathrm{~d} x=\left.\frac{1}{2} \cdot \frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}\right|_{0} ^{1}=\frac{2}{3} \sqrt{2}-\frac{1}{3} . $$ （7）积分 $\displaystyle f(x)=\frac{1}{1+x^{2}}+\sqrt{1-x^{2}} \int_{0}^{1} f(x) \mathrm{d} x$ 得 $$ \int_{0}^{1} f(x) \mathrm{d} x=\int_{0}^{1} \frac{1}{1+x^{2}} \mathrm{~d} x+\int_{0}^{1} \sqrt{1-x^{2}} \mathrm{~d} x \cdot \int_{0}^{1} f(x) \mathrm{d} x $$ 又 $\displaystyle \int_{0}^{1} \frac{1}{1+x^{2}} \mathrm{~d} x=\left.\arctan x\right|_{0} ^{1}=\frac{\pi}{4}, \int_{0}^{1} \sqrt{1-x^{2}} \mathrm{~d} x \xlongequal{x=\sin t} \int_{0}^{\frac{\pi}{2}} \cos t \cdot \cos t \mathrm{~d} t=\frac{\pi}{4}$ ，于是 $\displaystyle \int_{0}^{1} f(x) \mathrm{d} x=\frac{\pi}{4-\pi}$ ． （8）由 $F(x) F^{\prime}(x)=\sin ^{2} 2 x$ ，积分得 $$ F^{2}(x)=F^{2}(0)+2 \int_{0}^{x} \sin ^{2} 2 t \mathrm{~d} t=1+\int_{0}^{x}(1-\cos 4 t) \mathrm{d} t=1+x-\frac{1}{4} \sin 4 x $$ 于是 $$ F(x)=\sqrt{1+x-\frac{1}{4} \sin 4 x} $$ 求导得 $$ f(x)=\frac{1-\cos 4 x}{2 \sqrt{1+x-\frac{1}{4} \sin 4 x}} $$ （9） $\displaystyle \int_{0}^{1} x^{2} f^{\prime \prime}(2 x) \mathrm{d} x \xlongequal{t=2 x} \frac{1}{8} \int_{0}^{2} t^{2} f^{\prime \prime}(t) \mathrm{d} t=\frac{1}{8} \int_{0}^{2} t^{2} \mathrm{~d} f^{\prime}(t)=\frac{1}{8}\left(\left.t^{2} f^{\prime}(t)\right|_{0} ^{2}-2 \int_{0}^{2} t f^{\prime}(t) \mathrm{d} t\right)=-\frac{1}{4} \int_{0}^{2} t f^{\prime}(t) \mathrm{d} t$ $$ =-\frac{1}{4} \int_{0}^{2} t \mathrm{~d} f(t)=-\frac{1}{4}\left(\left.t f(t)\right|_{0} ^{2}-\int_{0}^{2} f(t) \mathrm{d} t\right)=-\frac{1}{4}\left(1-\int_{0}^{2} f(t) \mathrm{d} t\right)=0 $$