自学内容网 自学内容网

【数学二】一元函数积分学-不定积分与定积分的计算-定积分的换元积分法、分部积分法

考试要求

1、理解原函数的概念,理解不定积分和定积分的概念.
2、掌握不定积分的基本公式,掌握不定积分和定积分的性质及定积分中值定理,掌握换元积分法与分部积分法.
3、会求有理函数、三角函数有理式和简单无理函数的积分.
4、理解积分上限的函数,会求它的导数,掌握牛顿一菜布尼茨公式.
5、了解反常积分的概念,会计算反常积分.
6、掌握用定积分表达和计算一些几何量与物理量(平面图形的面积、平面曲线的弧长、旋转体的体积及侧面积、平行截面面积为已知的立体体积、功、引力、压力、质心、形心等)及函数的平均值.

不定积分与定积分的计算
定积分的换元积分法

定理 f ( x ) f(x) f(x) [ a , b ] [a,b] [a,b]上连续; x = φ ( t ) x=\varphi(t) x=φ(t)满足条件: a = φ ( a ) , b = φ ( b ) a=\varphi(a),b=\varphi(b) a=φ(a),b=φ(b),并且当 t t t在以 α , β \alpha,\beta α,β为端点的闭区间 I I I上变动时, a ≤ φ ( t ) ≤ b , φ ′ ( t ) a\le \varphi(t)\le b,\varphi^{'}(t) aφ(t)b,φ(t)连续,则有定积分的换元积分公式: ∫ a b f ( x ) d x = ∫ α β f ( φ ( t ) ) φ ′ ( t ) d t \int_{a}^{b}f(x)dx=\int_{\alpha}^{\beta}f(\varphi(t))\varphi^{'}(t)dt abf(x)dx=αβf(φ(t))φ(t)dt
练习1 ∫ 0 a a 2 − x 2 d x = \int_{0}^{a}\sqrt{a^2-x^2}dx= 0aa2x2 dx=?

令 x = a sin ⁡ t , 由 x ∈ [ 0 , 1 ] ⇒ t ∈ [ 0 , π 2 ] ∫ 0 a a 2 − x 2 d x ⟹ x = a sin ⁡ t ∫ 0 π 2 a 2 − ( a sin ⁡ t ) 2 d ( a sin ⁡ t ) = ∫ 0 π 2 a cos ⁡ t . a cos ⁡ t d t = a 2 ∫ 0 π 2 1 + cos ⁡ 2 t 2 d t = a 2 2 [ t ∣ 0 π 2 + sin ⁡ 2 t 2 ∣ 0 π 2 ] = π a 2 4 令x=a\sin t ,由x\in[0,1]\Rightarrow t\in[0,\frac{\pi}{2}] \\ \quad \\ \int_{0}^{a}\sqrt{a^2-x^2}dx \stackrel{x=a\sin t}\Longrightarrow \int_{0}^{\frac{\pi}{2}}\sqrt{a^2-(a\sin t )^2}d(a\sin t ) \\ \quad \\ =\int_{0}^{\frac{\pi}{2}}a\cos t.a\cos t dt\\ \quad \\ =a^2\int_{0}^{\frac{\pi}{2}}\frac{1+\cos 2t}{2} dt\\ \quad \\ =\frac{a^2}{2}[t|_{0}^{\frac{\pi}{2}}+\frac{\sin 2t}{2}|_{0}^{\frac{\pi}{2}}]\\ \quad \\ =\frac{\pi a^2}{4}\quad\quad\quad\quad\quad\quad x=asint,x[0,1]t[0,2π]0aa2x2 dxx=asint02πa2(asint)2 d(asint)=02πacost.acostdt=a202π21+cos2tdt=2a2[t02π+2sin2t02π]=4πa2

练习2:若 f ( x ) f(x) f(x)为连续函数,则 d x d ( ∫ 0 x f ( x − t ) d t ) = \frac{dx}{d}(\int_{0}^{x}f(x-t)dt)= ddx(0xf(xt)dt)=?

令 ξ = x − t ,由 t ∈ [ 0 , x ] ⇒ ξ ∈ [ x , 0 ] ∫ 0 x f ( x − t ) d t ⟹ ξ = x − t ∫ x 0 f ( ξ ) d ( x − ξ ) = − ∫ x 0 f ( ξ ) d ξ = ∫ 0 x f ( ξ ) d ξ ⇒ d x d ( ∫ 0 x f ( x − t ) d t ) = f ( x ) 令\xi=x-t,由t\in[0,x] \Rightarrow \xi\in[x,0] \\ \quad \\ \int_{0}^{x}f(x-t)dt \stackrel{\xi=x-t}\Longrightarrow \int_{x}^{0}f(\xi)d(x-\xi)=- \int_{x}^{0}f(\xi)d\xi \\ \quad \\ =\int_{0}^{x}f(\xi)d\xi \\ \quad \\ \Rightarrow \frac{dx}{d}(\int_{0}^{x}f(x-t)dt)=f(x) ξ=xt,由t[0,x]ξ[x,0]0xf(xt)dtξ=xtx0f(ξ)d(xξ)=x0f(ξ)dξ=0xf(ξ)dξddx(0xf(xt)dt)=f(x)

分部积分法

定理 u ( x ) , v ( x ) u(x),v(x) u(x),v(x)均有连续导数,则 ∫ u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) − ∫ v ( x ) u ′ ( x ) d x ∫ a b u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) ∣ a b − ∫ a b v ( x ) u ′ ( x ) d x \int u(x)v^{'}(x)dx=u(x)v(x)-\int v(x)u^{'}(x)dx \\ \quad \\ \int_{a}^{b} u(x)v^{'}(x)dx=u(x)v(x)|_{a}^{b}-\int_{a}^{b} v(x)u^{'}(x)dx u(x)v(x)dx=u(x)v(x)v(x)u(x)dxabu(x)v(x)dx=u(x)v(x)ababv(x)u(x)dx

练习1 ∫ ln ⁡ x d x \int \ln x dx lnxdx

解题1 ∫ ln ⁡ x d x = x ln ⁡ x − ∫ x d ln ⁡ x = x ln ⁡ x − x + C \int \ln x dx=x\ln x -\int xd \ln x=x\ln x-x+C lnxdx=xlnxxdlnx=xlnxx+C

练习2 ∫ x 2 sin ⁡ x d x \int x^2\sin xdx x2sinxdx

解题2 ∫ x 2 sin ⁡ x d x = x 2 . ( − cos ⁡ x ) − ∫ ( − cos ⁡ x ) 2 x d x = 2 ∫ x cos ⁡ x d x − x 2 cos ⁡ x = 2 [ x sin ⁡ x − ∫ sin ⁡ x d x ] − x 2 cos ⁡ x = 2 x sin ⁡ x + 2 cos ⁡ x − x 2 cos ⁡ x + C \int x^2\sin xdx=x^2.(-\cos x)-\int (-\cos x)2xdx \\ \quad \\=2\int x\cos x dx-x^2\cos x \\ \quad \\ =2[x\sin x-\int \sin xdx] -x^2\cos x \\ \quad \\ =2x\sin x+2\cos x-x^2\cos x+C x2sinxdx=x2.(cosx)(cosx)2xdx=2xcosxdxx2cosx=2[xsinxsinxdx]x2cosx=2xsinx+2cosxx2cosx+C

练习3 ∫ 0 π 2 e x sin ⁡ x d x \int_{0}^{\frac{\pi}{2}}e^x\sin xdx 02πexsinxdx

解题3 ∫ 0 π 2 e x sin ⁡ x d x = e x sin ⁡ x ∣ 0 π 2 − ∫ 0 π 2 e x cos ⁡ x d x = e x sin ⁡ x ∣ 0 π 2 − [ e x cos ⁡ x ∣ 0 π 2 + ∫ 0 π 2 e x sin ⁡ x d x ] = e π 2 + 1 − ∫ 0 π 2 e x sin ⁡ x d x ∫ 0 π 2 e x sin ⁡ x d x = e π 2 + 1 − ∫ 0 π 2 e x sin ⁡ x d x 合并同类项计算得: ∫ 0 π 2 e x sin ⁡ x d x = e π 2 + 1 2 \int_{0}^{\frac{\pi}{2}}e^x\sin xdx=e^x\sin x|_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}e^x\cos xdx \\ \quad \\ =e^x\sin x|_{0}^{\frac{\pi}{2}}-[e^x\cos x|_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}e^x\sin xdx] \\ \quad \\ =e^{\frac{\pi}{2}}+1-\int_{0}^{\frac{\pi}{2}}e^x\sin xdx \\ \quad \\ \int_{0}^{\frac{\pi}{2}}e^x\sin xdx=e^{\frac{\pi}{2}}+1-\int_{0}^{\frac{\pi}{2}}e^x\sin xdx \\ \quad \\ 合并同类项计算得:\int_{0}^{\frac{\pi}{2}}e^x\sin xdx=\frac{e^{\frac{\pi}{2}}+1}{2} 02πexsinxdx=exsinx02π02πexcosxdx=exsinx02π[excosx02π+02πexsinxdx]=e2π+102πexsinxdx02πexsinxdx=e2π+102πexsinxdx合并同类项计算得:02πexsinxdx=2e2π+1

练习4 ∫ 0 1 x e x 1 + e x d x \int_{0}^1\frac{xe^x}{\sqrt{1+e^x}}dx 011+ex xexdx

解题4 ∫ 0 1 x e x 1 + e x d x = ∫ 0 1 x d ( 2 1 + e x ) = 2 x 1 + e x ∣ 0 1 − ∫ 0 1 2 1 + e x d x ⟹ 1 + e x = t = 2 1 + e − 2 ∫ 2 1 + e t d ( ln ⁡ ( t 2 − 1 ) ) = 2 1 + e − 2 ∫ 2 1 + e [ 2 + 2 t 2 − 1 ] d t = 4 2 − 2 1 + e − 2 ∫ 2 1 + e [ 2 t 2 − 1 ] d t = 4 2 − 2 1 + e − 2 ∫ 2 1 + e [ 1 t − 1 − 1 t + 1 ] d t = 4 2 − 2 1 + e − 2 ln ⁡ t − 1 t + 1 ∣ 2 1 + e = 4 2 − 2 1 + e − 2 ln ⁡ 1 + e − 1 1 + e + 1 + 2 ln ⁡ 2 − 1 2 + 1 = 4 2 − 2 1 + e − 4 ln ⁡ ( 1 + e − 1 ) + 2 + 4 ln ⁡ ( 2 − 1 ) \int_{0}^1\frac{xe^x}{\sqrt{1+e^x}}dx=\int_{0}^1xd(2\sqrt{1+e^x})\\ \quad \\ =2x\sqrt{1+e^x}|_{0}^1-\int_{0}^12\sqrt{1+e^x}dx \\ \quad \\ \stackrel{\sqrt{1+e^x}=t}\Longrightarrow=2\sqrt{1+e}-2\int_{\sqrt{2}}^{\sqrt{1+e}}td(\ln(t^2-1))\\ \quad \\ =2\sqrt{1+e}-2\int_{\sqrt{2}}^{\sqrt{1+e}}[2+\frac{2}{t^2-1}]dt\\ \quad \\ =4\sqrt{2}-2\sqrt{1+e}-2\int_{\sqrt{2}}^{\sqrt{1+e}}[\frac{2}{t^2-1}]dt \\ \quad \\ =4\sqrt{2}-2\sqrt{1+e}-2\int_{\sqrt{2}}^{\sqrt{1+e}}[\frac{1}{t-1}-\frac{1}{t+1}]dt\\ \quad \\ =4\sqrt{2}-2\sqrt{1+e}-2\ln \frac{t-1}{t+1}|_{\sqrt{2}}^{\sqrt{1+e}}\\ \quad \\ =4\sqrt{2}-2\sqrt{1+e}-2\ln \frac{\sqrt{1+e}-1}{\sqrt{1+e}+1}+2\ln \frac{\sqrt{2}-1}{\sqrt{2}+1}\\ \quad \\ =4\sqrt{2}-2\sqrt{1+e}-4\ln(\sqrt{1+e}-1)+2+4\ln(\sqrt{2}-1) 011+ex xexdx=01xd(21+ex )=2x1+ex 010121+ex dx1+ex =t=21+e 22 1+e td(ln(t21))=21+e 22 1+e [2+t212]dt=42 21+e 22 1+e [t212]dt=42 21+e 22 1+e [t11t+11]dt=42 21+e 2lnt+1t12 1+e =42 21+e 2ln1+e +11+e 1+2ln2 +12 1=42 21+e 4ln(1+e 1)+2+4ln(2 1)

练习5 ∫ arcsin ⁡ e x e x d x \int \frac{\arcsin e^x}{e^x}dx exarcsinexdx

令 arcsin ⁡ e x = t , x = ln ⁡ ( sin ⁡ t ) ∫ t e ln ⁡ ( sin ⁡ t ) d ( ln ⁡ ( sin ⁡ t ) ) = ∫ t cos ⁡ t sin ⁡ 2 t d t = − ∫ t d 1 sin ⁡ t = − t sin ⁡ t + ∫ 1 sin ⁡ t d t = ∫ csc ⁡ x d x − t sin ⁡ t = ln ⁡ ∣ csc ⁡ t − cot ⁡ t ∣ − t sin ⁡ t + C = − e − x arcsin ⁡ e x + ln ⁡ ∣ e − x − e − 2 x − 1 ∣ + C 令\arcsin e^x=t,x=\ln(\sin t) \\ \quad \\ \int \frac{t}{e^{\ln(\sin t)}}d(\ln(\sin t))= \int \frac{t\cos t}{\sin^2 t}dt \\ \quad \\=-\int td\frac{1}{\sin t}=-\frac{t}{\sin t}+\int \frac{1}{\sin t}dt=\int \csc x dx-\frac{t}{\sin t}\\ \quad \\=\ln|\csc t - \cot t| -\frac{t}{\sin t}+C\\ \quad \\ =-e^{-x}\arcsin e^x+\ln|e^{-x} -\sqrt{e^{-2x}-1}|+C arcsinex=t,x=ln(sint)eln(sint)td(ln(sint))=sin2ttcostdt=tdsint1=sintt+sint1dt=cscxdxsintt=lncsctcottsintt+C=exarcsinex+lnexe2x1 +C

练习6 G ′ ( x ) = arcsin ⁡ ( x − 1 ) 2 , G ( 0 ) = 0 , 求 ∫ 0 1 G ( x ) d x G^{'}(x)=\arcsin (x-1)^2,G(0)=0,求\int_0^1G(x)dx G(x)=arcsin(x1)2G(0)=0,01G(x)dx

∫ 0 1 G ( x ) d x = G ( x ) . x ∣ 0 1 − ∫ 0 1 x d G ( x ) = G ( 1 ) − ∫ 0 1 x G ′ ( x ) d x = ∫ 0 1 G ′ ( x ) d x − ∫ 0 1 x G ′ ( x ) d x = ∫ 0 1 ( G ′ ( x ) − x G ′ ( x ) ) d x = − ∫ 0 1 ( x − 1 ) arcsin ⁡ ( x − 1 ) 2 d x = − 1 2 ∫ 0 1 arcsin ⁡ ( x − 1 ) 2 d ( x − 1 ) 2 ⟹ t = ( x − 1 ) 2 − 1 2 ∫ 1 0 arcsin ⁡ t d t = 1 2 ∫ 0 1 arcsin ⁡ t d t = 1 2 [ t . arcsin ⁡ t ∣ 0 1 − ∫ 0 1 t 1 − t 2 d t ] = 1 2 [ t . arcsin ⁡ t ∣ 0 1 + ∫ 0 1 d ( 1 − t 2 ) ] = 1 2 [ t . arcsin ⁡ t ∣ 0 1 + 1 − t 2 ∣ 0 1 ] = π 4 − 1 2 \int_0^1G(x)dx=G(x).x|_0^1-\int_0^1xdG(x) \\ \quad \\ =G(1)-\int_0^1xG^{'}(x)dx\\ \quad \\=\int_0^1G^{'}(x)dx-\int_0^1xG^{'}(x)dx\\ \quad \\ =\int_0^1(G^{'}(x)-xG^{'}(x)) dx \\ \quad \\ =-\int_0^1(x-1)\arcsin (x-1)^2dx\\ \quad \\ =-\frac{1}{2}\int_0^1\arcsin (x-1)^2d(x-1)^2 \\ \quad \\ \stackrel{t=(x-1)^2}\Longrightarrow -\frac{1}{2}\int_1^0\arcsin t dt=\frac{1}{2}\int_0^1\arcsin t dt \\ \quad \\ =\frac{1}{2}[t.\arcsin t|_0^1-\int_0^1\frac{t}{\sqrt{1-t^2}}dt] \\ \quad \\ =\frac{1}{2}[t.\arcsin t|_0^1+\int_0^1d(\sqrt{1-t^2})] \\ \quad \\ =\frac{1}{2}[t.\arcsin t|_0^1+\sqrt{1-t^2}|_0^1]=\frac{\pi}{4}-\frac{1}{2} 01G(x)dx=G(x).x0101xdG(x)=G(1)01xG(x)dx=01G(x)dx01xG(x)dx=01(G(x)xG(x))dx=01(x1)arcsin(x1)2dx=2101arcsin(x1)2d(x1)2t=(x1)22110arcsintdt=2101arcsintdt=21[t.arcsint01011t2 tdt]=21[t.arcsint01+01d(1t2 )]=21[t.arcsint01+1t2 01]=4π21


原文地址:https://blog.csdn.net/henni_719/article/details/142951126

免责声明:本站文章内容转载自网络资源,如本站内容侵犯了原著者的合法权益,可联系本站删除。更多内容请关注自学内容网(zxcms.com)!