第五章总复习题课

一、填空题

1.\begin{aligned} \lim_{x \to 0}\frac{\int_0^x{\rm cos}^2t}{x}{\rm d}t =\text[ \ 1\ ] \end{aligned}

解析：\begin{aligned} \lim_{x \to 0}\frac{\int_0^x{\rm cos}^2t}{x}{\rm d}t &=\lim_{x \to 0}\frac{(\int _0^x{\rm cos}^2t{\rm d}t)'}{x'}\\ &= \lim _{x \to 0}\frac{{\rm cos}^2x}{1}\\ &=1. \end{aligned}

2.\begin{aligned} \lim_{x \to 0}\frac{(\int _0^x{\rm e}^{t^2}{\rm d}t)^2}{\int_0^xt{\rm e}^{2t^2}{\rm d}t}=[\ 2\ ] \end{aligned}

解析：\begin{aligned} \lim_{x \to 0}\frac{(\int_0^x{\rm e}^{t^2}{\rm d}t)^2}{\int_0^xt{\rm e}^{2t^2}{\rm d}t} &=\lim_{x \to 0}\frac{2{\rm e}^{x^2}\int_0^x{\rm e}^{t^2}{\rm d}t}{x{\rm e}^{2x^2}}\\ &=\lim_{x \to 0}\frac{2\int_0^x{\rm e}^{t^2}{\rm d}t}{x{\rm e}^{x^2}}\\ &=\lim_{x \to 0}\frac{2{\rm e}^{x^2}}{{\rm e}^{x^2}+x \cdot {\rm e}^{x^2} \cdot 2x}\\ &=\lim_{x \to 0}\frac{2}{1+2x^2}\\ &=2. \end{aligned}

3.\begin{aligned} \int _{- \sqrt 2}^ {\sqrt 2}\frac{x}{(1+x^2)^2}{\rm d}x=[\ 0\ ] \end{aligned}

解析：t=1+x^2,{\rm d}t=2{\rm d}x.\\

\begin{aligned} \int _{- \sqrt 2}^ {\sqrt 2}\frac{x}{(1+x^2)^2}{\rm d}x &=\int_{-\sqrt2}^0\frac{x}{(1+x^2)^2}{\rm d}x+\int_0^{\sqrt2}\frac x{(1+x^2)^2}{\rm d}x\\ &=\frac12(\int_3^1\frac1{t^2}{\rm d}t+\int_1^3\frac1{t^2}{\rm d}t)\\ &=\left.\frac12(-\frac1t \right| _3^1-\left. \frac1t \right|_1^3)\\ &=0. \end{aligned}

4.\begin{aligned} \int _{-1}^1 x \mid x\mid {\rm d}x=[\ 0\ ] \end{aligned}

解析：由于y=x\mid x \mid是[-1,1]上的奇函数，所以\begin{aligned} \int _{-1}^1 x \mid x\mid {\rm d}x=[\ 0\ ]. \end{aligned}

二、选择题

5.由曲线\begin{aligned} y=\frac1x \end{aligned}与直线x=1,x=4,y=0所围成的平面图形的面积是[ A ].

A.2\ln2

B.2

C.\begin{aligned} \frac{\ln 2}{2} \end{aligned}

D.\ln 2

解析：\begin{aligned} \int_1^4(\frac1x-0){\rm d}x &=\int_1^4\frac1x{\rm d}x\\ &=[\ \ln \mid x \mid]_1^4\\ &=\ln 4-\ln 1\\ &=\ln 4=2\ln 2 \end{aligned}

6.求由曲线\begin{aligned} x^2=2y \end{aligned}与\begin{aligned} y^2=2x \end{aligned}所围成的图形的面积[ D ].

A.1

B.2

C.\begin{aligned} \frac 23 \end{aligned}

D.\begin{aligned} \frac 43 \end{aligned}

解析：\begin{aligned} S &= \int_0^2 \left( \sqrt{2x} - \frac{x^2}{2} \right) \mathrm{d}x \\ &= \left. \frac{2\sqrt{2}}{3} x^{\frac{3}{2}} \right|_{0}^{2}-\left.\frac{x^3}6\right|_0^2\\ &=\frac83-\frac86\\ &=\frac 43. \end{aligned}

7.计算定积分\begin{aligned} \int_0^{\frac \pi2}{\rm sin}\ x{\rm cos}^3x{\rm d}x= \end{aligned}[ D ].

A.1

B.\begin{aligned} \frac12 \end{aligned}

C.\begin{aligned} \frac13 \end{aligned}

D.\begin{aligned} \frac14 \end{aligned}

解析：\begin{aligned} \int_0^{\frac \pi2}{\rm sin}\ x{\rm cos}^3x{\rm d}x &=-\int_0^{\frac \pi2}{\rm cos}^3x{\rm dcos}\ x\\ &=\left.-\frac14[{\rm cos}^4x]\right|_0^{\frac\pi2}\\ &=\frac14. \end{aligned}

8.计算定积分\begin{aligned} \int_{-2}^2\frac{x+\mid x\mid}{2+x^2}{\rm d}x= \end{aligned}[ C ].

A.1

B.0

C.\ln3

D.\ln6

解析：\begin{aligned} \int_{-2}^2\frac{x+\mid x\mid}{2+x^2}{\rm d}x &=\int_{-2}^0\frac{x+\mid x\mid }{2+x^2}{\rm d}x+\int_0^2\frac{x+\mid x\mid}{2+x^2}{\rm d}x\\ &=\int_0^2\frac{2x}{2+x^2}{\rm d}x\\ &=\int_0^2\frac1{2+x^2}{\rm d}(2+x^2)\\ &=\left.\ln(2+x^2)\right|_0^2\\ &=\ln6-\ln 2\\ &=\ln\frac62=\ln3. \end{aligned}

三、计算题

9.计算定积分\begin{aligned} \int_1^ex\ln x{\rm d}x. \end{aligned}

解答：\begin{aligned} &u=\ln x,{\rm d}v=x{\rm d}x={\rm d}(\frac {x^2}{2})\\ &{\rm d}u=\frac1x{\rm d}x,v=\frac12x^2 \end{aligned}

\begin{aligned} \int_1^ex\ln x{\rm d}x &=\int_1^e\frac{\ln x}{2}{\rm d}(x^2)\\ &=\left.[\frac12x^2\ln x]\right|_1^e-\int_1^e\frac x2{\rm d}x\\ &=\frac{{\rm e}^2+1}4. \end{aligned}

10.计算定积分\begin{aligned} \int_0^1 x{\rm arctan \ }x{\rm d}x. \end{aligned}

解答：\begin{aligned} &u={\rm arctan\ }x,{\rm d}v=x{\rm d}x={\rm d}(\frac {x^2}{2})\\ &{\rm d}u=\frac1{1+x^2}{\rm d}x,v=\frac12x^2 \end{aligned}

\begin{aligned} \int_0^1 x{\rm arctan \ }x{\rm d}x &=\frac12\int_0^1{\rm arctan\ }x{\rm d}(x^2)\\ &=\left.[\frac 12x^2{\rm arctan\ }x]\right|_0^1-\frac12\int_0^1\frac{x^2}{1+x^2}{\rm d}x\\ &=\frac 12*1*\frac{\pi}{4}-\frac12\int_0^1(1-\frac1{1+x^2}){\rm d}x\\ &=\frac{\pi}{8}-\left. \frac12[x-{\rm arctan\ }x]\right|_0^1\\ &=\frac{\pi}{8}-\frac12(1-\frac{\pi}{4})\\ &=\frac{\pi}4-\frac12. \end{aligned}

四、证明题

11.证明\begin{aligned} \int_0^{\frac\pi4}{\rm sin^4}x{\rm d}x< \int_0^{\frac\pi4}{\rm sin^2}x{\rm d}x. \end{aligned}

解答：因为在积分区间[0,{\frac \pi 4}]上，0 \leq {\rm sin}\ x<1.所以{\rm sin^4}x<{\rm sin^2}x.

故有\begin{aligned} \int_0^{\frac\pi4}{\rm sin^4}x{\rm d}x< \int_0^{\frac\pi4}{\rm sin^2}x{\rm d}x. \end{aligned}

12.证明\begin{aligned} \int_e^{2e}\ln x\ {\rm d}x< \int_e^{2e}(\ln x)^2{\rm d}x. \end{aligned}

解答：因为在积分区间[{\rm e},{\rm 2e}]上，\ln x \geq1.所以\ln x<(\ln x)^2.

故有\begin{aligned} \int_e^{2e}\ln x\ {\rm d}x< \int_e^{2e}(\ln x)^2{\rm d}x. \end{aligned}

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