高等数学(积分学)

一. 不定积分的概念

1.1 原函数

$(x^2{)}^{\prime }=?$ \quad \quad $(?{)}^{\prime }=2x$
$(原函数{)}^{\prime }=导数$

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思考一下$(?{)}^{\prime }=2x$
有 $x^2,x^2+1,x^2-6,x^2-e...$
我们用$x^2+C$来代表全体原函数

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1.2 不定积分

$F^{\prime }(x)=f(x)$

$\int f(x)dx=F(x)+C$ \quad (不定积分)
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例题1: $\int \sin xdx$
$(-\cos x{)}^{\prime }=\sin x$
$\int \sin xdx=-\cos x+C$

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例题2: $\int e^{x}dx=e^x+C$

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例题3: $\int dx=\int 1dx=x+C$

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例题4: $\int 0dx=C$

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例题5: $\int \sqrt{x\sqrt{x}}dx=x^{\frac{3}{4}}+C$

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二. 直接积分法

2.1 法则

加减: $\int (\sin x\pm e^x)dx=\int \sin xdx\pm \int e^{x}dx$
数乘: $\int c\sin xdx=c\int \sin xdx$ \quad (常数可以提出来)

特别注意: 积分不能分开乘除
如:
$\int \sin x\ast \cos xdx\ne \int \sin xdx\ast \int \cos xdx$
$\int x\cos xdx\ne x\int \cos xdx$ \quad (x不能提出来)
$\int \frac{\cos x}{x}dx\ne \int \frac{\int \cos xdx}{\int xdx}$

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2.2 公式

$\int x^{3}dx=\frac{1}{4}{x}^4+C$

$\int 3^{x}dx=\frac{3^x}{\ln 3}+C$

$\int \frac{1}{x}dx=\ln |x|+C$

$\int {\sec }^{2}xdx=\tan x+C$

$\int cs{c}^{2}xdx=-\cot x+C$

$\int \tan x\sec xdx=\sec x+C$

$\int \cot x\csc dx=-\csc x+C$

$\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin x+C$

$\int \frac{-1}{\sqrt{1-x^2}}dx=\arccos x+C$

$\int \frac{1}{1+x^2}dx=\arctan x+C$

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例题6: $\int \frac{(x+1{)}^2}{x}dx$
=$\int \frac{x^2+2x+1}{x}dx$
=$\int x+2+\frac{1}{x}dx$
=$\frac{1}{2}{x}^2+2x+\ln |x|+C$

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例题7: $\int \frac{x^2+\sqrt{x}{e}^x+1}{\sqrt{x}}dx$
=$\int \frac{x^2}{\sqrt{x}}+e^x+\frac{1}{\sqrt{x}}dx$
=$\int x\sqrt{x}+e^x+\frac{1}{\sqrt{x}}dx$
=$\int x^{\frac{3}{2}}+e^x+x^{-\frac{1}{2}}dx$
=$\frac{2}{5}{x}^{\frac{5}{2}}+e^x+2x^{\frac{1}{2}}+C$

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例题8: $\int 3^x{e}^{x}dx$
=$\int (3e{)}^{x}dx$
=$\frac{(3e{)}^x}{\ln 3e}+C$

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例题9: $\int \frac{(x+1{)}^2}{x(1+x^2)}dx$
=$\int \frac{x^2+1}{x(1+x^2)}+\frac{2x}{x(1+x^2)}dx$

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例题10: $\int \frac{1+3x^2}{x^2(1+x^2)}dx$
=$\int \frac{3x^2}{x^2(1+x^2)}+\frac{1}{x^2(1+x^2)}dx$
=$\int \frac{3}{(1+x^2)}+\frac{1+x^2-x^2}{x^2(1+x^2)}dx$
=$\int \frac{3}{(1+x^2)}+\frac{1+x^2}{x^2(1+x^2)}-\frac{x^2}{x^2(1+x^2)}dx$
=$3\arctan x-x^{-1}-\arctan x+C$
=$2\arctan x-x^{-1}+C$

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例题11: $f(x)$的一个原函数是$\ln x$,求$f^{\prime }(x)$
依题意得: $(\ln x{)}^{\prime }=f(x)$
$\therefore$ $f(x)=\frac{1}{x}$
$\therefore$ $f^{\prime }(x)=-\frac{1}{x^2}$

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三. 第一类换元法(凑微分)

推导:
$\int \cos xdx=\sin x+C$
-> $\int \cos udu=\sin u+C$
-> $\int \cos 3xd3x=\sin 3x+C$

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例题12: $\int e^{-x}dx$
=$-\int e^{-x}d(-x)$
=$-e^{-x}+C$

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例题13: $\int \cos (3x+2)dx$
=$\frac{1}{3}\int \cos (3x+2)d3x$
=$\frac{1}{3}\int \cos (3x+2)d(3x+2)$ \quad \quad (后面可以加减常数,结果不变)
=$\frac{1}{3}\sin (3x+2)+C$

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例题14: $\int e^{1-x}dx$
=$-\int e^{1-x}d(-x)$
=$-\int e^{1-x}d(-x+1)$
=$-e^{1-x}+C$

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例题15: $\int \sin (2x-1)dx$
=$\frac{1}{2}\int \sin (2x-1)d2x$
=$\frac{1}{2}\int \sin (2x-1)d(2x-1)$
=$-\frac{1}{2}\cos (2x-1)+C$

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例题16: $\int \sqrt{2x+1}dx$
=$\frac{1}{2}\int \sqrt{2x+1}d2x$
=$\frac{1}{2}\int \sqrt{2x+1}d(2x+1)$
=$\frac{1}{2}\ast \frac{2}{3}(2x+1{)}^{\frac{3}{2}}+C$
=$\frac{1}{3}(2x+1{)}^{\frac{3}{2}}+C$

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例题17: $\int \frac{1}{1+4x^2}dx$
=$\int \frac{1}{1+(2x{)}^2}dx$
=$\frac{1}{2}\int \frac{1}{1+(2x{)}^2}d(2x)$
=$\frac{1}{2}\arctan 2x+C$

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例题18: $\int \frac{1}{2x+1}dx$
=$\frac{1}{2}\int \frac{1}{2x+1}d2x$
=$\frac{1}{2}\int \frac{1}{2x+1}d(2x+1)$
=$\frac{1}{2}\ln (2x+1)+C$

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例题19: $\int x{e}^{x^2}dx$
=$\frac{1}{2}\int e^{x^2}2xdx$
=$\frac{1}{2}\int e^{x^2}(x^2{)}^{\prime }dx$
=$\frac{1}{2}\int e^{x^2}d{x}^2$
=$\frac{1}{2}{e}^{x^2}+C$

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例题20: $\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx$
=$\int e^{\sqrt{x}}\ast \frac{1}{\sqrt{x}}dx$
=$\int e^{\sqrt{x}}\ast (2\sqrt{x}{)}^{\prime }dx$
=$\int e^{\sqrt{x}}d(2\sqrt{x})$
=$2\int e^{\sqrt{x}}d\sqrt{x}$
=$2e^{\sqrt{x}}+C$

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例题21: $\int \frac{\ln x}{x}dx$
=$\int \ln x\ast \frac{1}{x}dx$
=$\int \ln x\ast (\ln x{)}^{\prime }dx$
=$\int \ln xd(\ln x)$ 令$\ln x$为$u$
=$\int udu$
=$\frac{1}{2}{u}^2+C$
=$\frac{1}{2}(\ln x{)}^2+C$

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例题22: $\int \frac{2x}{2+x^2}dx$
=$\int \frac{1}{2+x^2}\ast 2xdx$
=$\int \frac{1}{2+x^2}(x^2{)}^{\prime }dx$
=$\int \frac{1}{2+x^2}d(x^2+2)$
=$\ln (x^2+2)+C$

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例题23: 若$\int f(x)dx=e^{2x}+C$,则$f(x)=$_____
$2e^{2x}$

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例题24:

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例题25: $\int x\sqrt{1-x^2}dx$
=$\int \sqrt{1-x^2}xdx$
=$\frac{1}{2}\int \sqrt{1-x^2}2xdx$
=$\frac{1}{2}\int \sqrt{1-x^2}(x^2{)}^{\prime }dx$
=$-\frac{1}{2}\int \sqrt{1-x^2}d(-x^2+1)$
=$-\frac{1}{2}\ast \frac{2}{3}(1-x^2{)}^{\frac{3}{2}}+C$
=$-\frac{1}{3}(1-x^2{)}^{\frac{3}{2}}+C$

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理解:

1. $[\int f(x)dx{]}^{\prime }=f(x)$

2. $d\int f(x)dx=f(x)dx$

3. $\int F^{\prime }(x)dx=F(x)+C$

4. $\int dF(x)=F(x)+C$ \quad $\int d\sin \sqrt{x}=\sin \sqrt{x}+C$

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例题26: $\int {\cos }^{3}xdx$
=$\int {\cos }^{2}x\ast \cos xdx$
=$\int (1-{\sin }^{2}x)(\sin x{)}^{\prime }dx$
=$\int (1-{\sin }^{2}x)d\sin x$
=$\int 1d\sin x-\int {\sin }^{2}xd\sin x$
=$\sin x-\frac{1}{3}(\sin x{)}^3+C$

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四. 第二类换元法

实在没办法了且带根号的才用第二换元法

第一类根换

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例题27: $\int \frac{1}{1+\sqrt{x}}dx$ \quad \quad \quad 令$\sqrt{x}=u,x=u^2$
=$\int \frac{1}{1+u}d{u}^2$
=$\int \frac{1}{1+u}(u^2{)}^{\prime }du$
=$\int \frac{2u}{1+u}du$
=$2\int \frac{u}{1+u}du$
=$2\int \frac{u+1-1}{1+u}du$
=$2\int \frac{1+u}{1+u}-\frac{1}{1+u}du$
=$2\int 1du-\int \frac{1}{1+u}du$
=$2u-\ln (1+u)+C$
=$2\sqrt{x}-\ln (1+\sqrt{x})+C$

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例题28: $\int \frac{1}{\sqrt{1+x}}dx$ \quad \quad 令$\sqrt{1+x}=u,x=u^2-1$
=$\int \frac{1}{u}d(u^2-1)$
=$\int \frac{1}{u}d{u}^2$
=$\int \frac{1}{u}(u^2{)}^{\prime }du$
=$\int \frac{2u}{u}du$
=$\int 2du$
=$2u+C$
=$2\sqrt{1+x}+C$

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例题29: $\int x\sqrt{x+1}dx$ \quad \quad 令$\sqrt{x+1}=u,x=u^2-1$
=$\int u(u^2-1)d(u^2-1)$
=$\int u(u^2-1)(u^2{)}^{\prime }du$
=$\int u(u^2-1)2udu$
=$2\int u^4-u^{2}du$
=$2(\frac{1}{5}{u}^5-\frac{1}{3}{u}^3)+C$
=$\frac{2}{5}(\sqrt{x+1}{)}^5-\frac{2}{3}(\sqrt{x+1}{)}^3+C$

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五. 分部积分法

$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$
$u{v}^{\prime }=(uv{)}^{\prime }-u^{\prime }v$

$\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx$

打油诗
多指弦, $u$选多 \quad \quad (多项式, 指数, 三角函数)
多反对, $u$不选多 \quad \quad (多项式, 反函数, 对数)

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例题30: $\int x\ln xdx$ \quad \quad (多对)
=$\int \frac{1}{2}{x}^2\ln x-\int \frac{1}{x}\ast \frac{1}{2}{x}^{2}dx$
=$\int \frac{1}{2}{x}^2\ln x-\int \frac{1}{2}xdx$
=$\frac{1}{2}{x}^2\ln x-\frac{1}{4}{x}^2+C$

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例题31: $\int x{e}^{-x}dx$ \quad \quad (多指) $u=x,u^{\prime }=1,v^{\prime }=e^{-x},v=-e^{-x}$
=$-x{e}^{-x}-\int -e^{-x}dx$
=$-x{e}^{-x}+\int e^{-x}dx$
=$-x{e}^{-x}-e^{-x}+C$

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例题32: $\int \ln xdx$
=$x\ln x-\int \frac{1}{x}\ast xdx$
=$x\ln x-\int 1dx$
=$x\ln x-x+C$

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例题33: $\int x\arctan xdx$
=$\frac{1}{2}{x}^2\arctan x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx$
=$\frac{1}{2}{x}^2\arctan x-\frac{1}{2}\int \frac{x^2+1-1}{x^2+1}dx$
=$\frac{1}{2}{x}^2\arctan x-\frac{1}{2}(\int \frac{x^2+1}{x^2+1}dx-\int \frac{1}{x^2+1}dx)$
=$\frac{1}{2}{x}^2\arctan x-\frac{1}{2}x+\frac{1}{2}\arctan x+C$

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例题34: $\int x\cos xdx$
=$x\sin x-\int \sin xdx$
=$x\sin x-(-\cos x)+C$
=$x\sin x+\cos x+C$

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例题35: $\int \ln (1+x)dx$
=$\int 1\ast \ln (1+x)dx$
=$x\ln (1+x)-\int \frac{x}{1+x}dx$
=$x\ln (1+x)-\int \frac{x+1-1}{1+x}dx$
=$x\ln (1+x)-(\int \frac{1+x}{1+x}dx-\int \frac{1}{1+x}dx)$
=$x\ln (1+x)-x+\ln (1+x)+C$

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六. 定积分

用于求曲形面积



面积只能为正, 积分有正有负

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6.1. 定积分的性质

1. ${\int }_a^{a}f(x)dx=0$

2. ${\int }_a^{b}dx={\int }_a^{b}1dx=b-a$

3. ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$

4. ${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$ \quad \quad 定积分的区间可加性

不论a,b,c的相对位置如何, 上式都成立

5. ${\int }_a^b[f(x)\pm g(x)]dx={\int }_a^{b}f(x)dx\pm {\int }_a^{b}g(x)dx$ \quad \quad 乘除不行

6. ${\int }_a^{b}k\ast f(x)dx=k\ast {\int }_a^{b}f(x)dx$

7. ${\int }_{-a}^{a}f(x)dx=0$ \quad \quad 奇函数在对称区间上的定积分为0

奇$\pm$奇=奇	奇(×)(÷)奇=偶
奇$\pm$偶 = 非奇非偶	奇(×)(÷)偶=奇
偶$\pm$(×)(÷)偶=偶

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例题36:
(1) ${\int }_{-1}^1\sin xdx=0$

(2) ${\int }_{-2}^{2}x\cos xdx=0$

(3) ${\int }_{-3}^3\frac{x}{\sqrt{1+x^2}}dx=0$

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例题37: ${\int }_{-1}^1(2\sin x+3)dx=$_________

${\int }_{-1}^{1}2\sin xdx+{\int }_{-1}^{1}3dx=0+3\ast 2=6$

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例题38: ${\int }_{-1}^1(5{\sin }^{2023}x-2\tan x+3)dx=$________
${\int }_{-1}^{1}5{\sin }^{2023}xdx-{\int }_{-1}^{1}2\tan xdx+{\int }_{-1}^{1}3dx=0-0+6$

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七. 牛顿-莱布尼兹公式(N-L)

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例题39: ${\int }_0^1{x}^{2}dx=$_____
$\frac{1}{3}{x}^3{|}_0^1=\frac{1}{3}-0=\frac{1}{3}$

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例题40: ${\int }_{-1}^{1}19x^{18}(1+\sin x)dx=$______
${\int }_{-1}^{1}19x^{18}dx+{\int }_{-1}^1\sin xdx=19\ast \frac{1}{19}{x}^{19}{|}_{-1}^1+0=1-(-1)=2$

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例题41:
$$f(x)=\{\begin{array}{ll}x& x\le 1\\ 2x& x>1\end{array}$$
求${\int }_0^{2}f(x)dx$

=${\int }_0^{1}f(x)dx+{\int }_1^{2}f(x)dx$
=$\frac{1}{2}{x}^2{|}_0^1+x^2{|}_1^2$
=$3\frac{1}{2}$

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例题42: ${\int }_0^1(2x-1{)}^{5}dx=$_______
=$\frac{1}{2}{\int }_0^1(2x-1{)}^{5}d(2x)$
=$\frac{1}{2}{\int }_0^1(2x-1{)}^{5}d(2x-1)$
=$\frac{1}{2}\ast \frac{1}{6}(2x-1{)}^6{|}_0^1$
=$0$

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例题43: ${\int }_{-2}^{-1}\frac{1}{1+\sqrt{2x+5}}dx$

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八. 定积分的分部积分法

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例题44: ${\int }_0^{\frac{\pi }{2}}x\sin xdx$

=$[0-0]-[-1-0]=1$

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例题45:

=$1$

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例题46:

=$\frac{e^2}{4}+\frac{1}{4}$

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例题47:

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例题48:

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例题49: ${\int }_{-1}^{2}x{e}^{x^2}dx$
=$\frac{1}{2}{\int }_{-1}^2{e}^{x^2}\ast 2xdx$
=$\frac{1}{2}{\int }_{-1}^2{e}^{x^2}\ast (x^2{)}^{\prime }dx$
=$\frac{1}{2}{\int }_{-1}^2{e}^{x^2}d(x^2)$
=$\frac{e^4}{2}-\frac{e}{2}$

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九. 定积分的几何应用

9.1 平面图形的面积

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例50: 计算由直线$y=x$和抛物线$y=x^2$所围成的平面图形的面积

交点(0,1), (1,1)

$S_{夹}={\int }_0^{1}xdx-{\int }_0^1{x}^{2}dx=\frac{1}{2}{x}^2{|}_0^1-\frac{1}{3}{x}^3{|}_0^1=\frac{1}{6}$

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例51: 计算由直线$y=x+2$和抛物线$y=x^2$所围成的平面图形的面积

$$\{\begin{array}{l}y=x+2\\ y=x^2\end{array}$$
解出交点为(-1,1), (2,4)

$S={\int }_{-1}^2(x+2)-x^{2}dx=(\frac{1}{2}{x}^2+2x-\frac{1}{3}{x}^3){|}_{-1}^2=\frac{9}{2}$

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9.2 旋转体的体积

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例52: 设曲线$y=e^x$, 与直线$x=0,x=1$和$x$轴围成的平面图形为D
(1) 求D的面积
(2) 求D围绕x轴旋转一周所得旋转体的体积

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例53: 一直平面图形D由曲线$y=e^x,y=x,x=0,x=1$围成
(1) 求D的面积A
(2) 求D绕x轴旋转一周所得的体积V

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例54: 设曲线$y=\cos x(x\in [0,\frac{\pi }{2}])$与x轴及y轴所围成的平面图形为D
(1) 求D的面积A
(1) 求D绕x轴旋转一周所得旋转体的体积

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例55: 已知平面图形D由曲线$y=\sqrt{1+x},y=1+x$围成
(1) 求D的面积A
(1) 求D绕x轴旋转一周所得旋转体的体积

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十. 广义积分

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例: ${\int }_0^1\frac{1}{\sqrt{x}}dx=$_______

${\int }_0^1{x}^{-\frac{1}{2}}dx=2x^{\frac{1}{2}}{|}_0^1=2$

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例: ${\int }_{-\infty }^1{e}^{x}dx=$______

$e^x{|}_{-\infty }^1=e-0=e$

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例: ${\int }_{-\infty }^{+\infty }\frac{1}{1+x^2}dx$______

$\arctan x{|}_{-\infty }^{+\infty }=\frac{\pi }{2}-(-\frac{\pi }{2})=\pi$

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十一. 变上限积分

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例: 设函数$f(x)={\int }_0^x{\cos }^{2}tdt$, 则$f^{\prime }(x)=$
${\cos }^{2}x$

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例: 设函数$f(x)={\int }_0^x\frac{1}{1+t^2}dt$, 则$f^{\prime }(x)=$
$\frac{1}{1+x^2}$