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التكامل بالأجزاء

أجد كلاً من التكاملات الآتية:

$\int (x+1)\cos xdx$ (1)

$\begin{array}{ll}u=x+1& dv=\cos xdx\\ du=dx& v=\sin x\\ \int (x+1)\cos xdx=& (x+1)\sin x-\int \sin xdx\\ & =(x+1)\sin x+\cos x+C\end{array}$

$\int x{e}^{x/2}dx$ (2)

$\begin{array}{rl}& u=xdv=e^{\frac{1}{2}x}dx\\ & du=dxv=2e^{\frac{1}{2}x}\\ & \int x{e}^{\frac{1}{2}x}dx=2x{e}^{\frac{1}{2^x}}-\int 2e^{\frac{1}{2}x}dx\\ & =2x{e}^{\frac{1}{2}x}-4e^{\frac{1}{2}x}+C\end{array}$

$\int (2x^2-1)e^{-x}dx$ (3)

$$\begin{gathered} \begin{array}{lrl}u=2x^2-1& dv& =e^{-x}dx\\ du=4xdx& v& =-e^{-x}\end{array} \\ \int (2x^2-1)e^{-x}dx=-(2x^2-1)e^{-x}+\int 4x{e}^{-x}dx \\ \begin{array}{ll}u=4x& dv=e^{-x}dx\\ du=4dx& v=-e^{-x}\end{array} \\ \begin{array}{rl}\int (2x^2-1)e^{-x}dx& =-(2x^2-1)e^{-x}-4x{e}^{-x}+\int 4e^{-x}dx\\ & =-(2x^2-1)e^{-x}-4x{e}^{-x}-4e^{-x}+C\\ & =-e^{-x}(2x^2+4x+3)+C\end{array} \end{gathered}$$

$\int \ln \sqrt{x}dx$ (4)

$\begin{array}{rl}& \int \ln \sqrt{x}dx=\int \frac{1}{2}\ln xdx\\ & u=\ln xdv=\frac{1}{2}dx\\ & du=\frac{1}{x}dxv=\frac{1}{2}x\\ & \int \frac{1}{2}\ln xdx=\frac{1}{2}x\ln x-\int \frac{1}{2}dx\\ & =\frac{1}{2}x\ln x-\frac{1}{2}x+C\end{array}$

$\int x\sin x\cos xdx$ (5)

$\begin{array}{rl}& \int x\sin x\cos xdx=\int \frac{1}{2}x\sin 2xdx\\ & u=\frac{1}{2}xdv=\sin 2xdx\\ & du=\frac{1}{2}dxv=-\frac{1}{2}\cos 2x\\ & \int x\sin x\cos xdx=-\frac{1}{4}x\cos 2x+\int \frac{1}{4}\cos 2xdx\\ & =-\frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x+C\end{array}$

$\int x\sec x\tan xdx$ (6)

$\begin{array}{rl}& u=xdv=\sec x\tan xdx\\ & du=dxv=\sec x\\ & \int x\sec x\tan xdx=x\sec x-\int \sec xdx\\ & =x\sec x-\int \sec x\times \frac{\sec x+\tan x}{\sec x+\tan x}dx\\ & =x\sec x-\int \frac{{\sec }^{2}x+\sec x\tan x}{\sec x+\tan x}dx\\ & =x\sec x-\ln |\sec x+\tan x|+C\end{array}$

$\int \frac{x}{{\sin }^{2}x}dx$ (7)

$\begin{array}{ll}\int \frac{x}{{\sin }^{2}x}dx=& \int x{\csc }^{2}xdx\\ u=x& dv={\csc }^{2}xdx\\ du=dx& v=-\cot x\\ \int x{\csc }^{2}xdx& =-x\cot x+\int \cot xdx\\ & =-x\cot x+\int \frac{\cos x}{\sin x}dx\\ & =-x\cot x+\ln |\sin x|+C\end{array}$

$\int \frac{\ln x}{x^3}dx$ (8)

$\begin{array}{rl}u=\ln x& dv=x^{-3}dx\\ du=\frac{1}{x}dx& v=-\frac{1}{2}{x}^{-2}\\ \int x^{-3}\ln xdx& =-\frac{1}{2}{x}^{-2}\ln x-\int -\frac{1}{2}{x}^{-2}\frac{1}{x}dx\\ & =-\frac{1}{2}{x}^{-2}\ln x+\int \frac{1}{2}{x}^{-3}dx\\ & =-\frac{1}{2}{x}^{-2}\ln x-\frac{1}{4}{x}^{-2}+C\\ & =-\frac{\ln x}{2x^2}-\frac{1}{4x^2}+C\end{array}$

$\int 2x^2{\sec }^{2}x\tan xdx$ (9)

$\begin{array}{ll}u=2x^2& dv={\sec }^{2}x\tan xdx\\ du=4xdx& v=\frac{1}{2}{\tan }^{2}x\end{array}$

ملاحظة: لإيجاد v استخدمنا طريقة التعويض، حيث: $y=\tan x,dx=\frac{dy}{{\sec }^{2}x}$ ومنه:

$$\begin{gathered} \begin{array}{rl}& v=\int se{c}^{2}x\tan xdx=\int se{c}^{2}xy\frac{dy}{se{c}^{2}x}=\int ydy=\frac{1}{2}{y}^2=\frac{1}{2}{\tan }^{2}x\\ & \int 2x^{2}se{c}^{2}x\tan xdx=2x^2(\frac{1}{2}{\tan }^{2}x)-\int 2x{\tan }^{2}xdx\end{array} \\ \begin{array}{ll}u=2x& dv={\tan }^{2}xdx=(se{c}^{2}x-1)dx\\ du=2dx& v=\tan x-x\\ \int 2x^{2}se{c}^{2}x\tan xdx& \end{array} \\ \begin{array}{rl}& =x^2{\tan }^{2}x-\left(2x\right(\tan x-x)-\int 2(\tan x-x\left)dx\right)\\ & =x^2{\tan }^{2}x-2x\tan x+2x^2+2\int (\frac{\sin x}{\cos x}-x)dx\\ =& x^2{\tan }^{2}x-2x\tan x+2x^2-2\ln |\cos x|-x^2+C\\ =& x^2{\tan }^{2}x-2x\tan x+x^2-2\ln |\cos x|+C\end{array} \end{gathered}$$

$\int (x-2)\sqrt{8-x}dx$ (10)

هذه المسألة يمكن حلها بالتعويض، حيث: $(u=\sqrt{8-x} \mathrm{أو} u=8-x)$ 

وحلها بالأجزاء كالآتي:

$\begin{array}{rl}& u=x-2dv=(8-x{)}^{\frac{1}{2}}dx\\ & du=dxv=-\frac{2}{3}(8-x{)}^{\frac{3}{2}}\\ & \int (x-2)\sqrt{8-x}dx=(x-2)\times -\frac{2}{3}(8-x{)}^{\frac{3}{2}}-\int -\frac{2}{3}(8-x{)}^{\frac{3}{2}}dx\\ & =-\frac{2}{3}(x-2)(8-x{)}^{\frac{3}{2}}-\frac{4}{15}(8-x{)}^{\frac{5}{2}}+C\end{array}$

$\int x^3\cos 2xdx$ (11)

بالأجزاء 3 مرات، لنستخدم طريقة الجدول:

$\int x^3\cos 2xdx=\frac{1}{2}{x}^3\sin 2x+\frac{3}{4}{x}^2\cos 2x-\frac{3}{4}x\sin 2x-\frac{3}{8}\cos 2x+C$

$\int \frac{x}{6^x}dx$ (12)

$\begin{array}{rl}& \int \frac{x}{6^x}dx=\int x{6}^{-x}dx\\ & u=xdv=6^{-x}dx\\ & du=dxv=-\frac{6^{-x}}{\ln 6}\\ & \int x{6}^{-x}dx=-x\frac{6^{-x}}{\ln 6}+\int \frac{6^{-x}}{\ln 6}dx\\ & =-x\frac{6^{-x}}{\ln 6}-\frac{6^{-x}}{(\ln 6{)}^2}+C\end{array}$

$\int e^{-x}\sin 2xdx$ (13)

$$\begin{gathered} \begin{array}{rl}& u=e^{-x}dv=sin2xdx\\ & du=-e^{-x}dxv=\frac{-1}{2}cos2x\\ & \int e^{-x}sin2xdx=-\frac{1}{2}{e}^{-x}cos2x-\int \frac{1}{2}{e}^{-x}cos2xdx\end{array} \\ \begin{array}{rl}& u=\frac{1}{2}{e}^{-x}dv=cos2xdx\\ & du=-\frac{1}{2}{e}^{-x}dxv=\frac{1}{2}sin2x\end{array} \\ \begin{array}{rl}& \int e^{-x}sin2xdx=-\frac{1}{2}{e}^{-x}cos2x-\frac{1}{4}{e}^{-x}sin2x-\frac{1}{4}\int e^{-x}sin2xdx\\ & \int e^{-x}sin2xdx+\frac{1}{4}\int e^{-x}sin2xdx=-\frac{1}{4}{e}^{-x}(sin2x+2cos2x)+C\\ & \frac{5}{4}\int e^{-x}sin2xdx=-\frac{1}{4}{e}^{-x}(sin2x+2cos2x)+C\\ & \int e^{-x}sin2xdx=-\frac{1}{5}{e}^{-x}(sin2x+2cos2x)+C\end{array} \end{gathered}$$

$\int \cos x\ln \sin xdx$ (14)

$$\begin{gathered} \begin{array}{ll}u=lnsinx& dv=cosxdx\\ du=\frac{cosx}{sinx}dx& v=sinx\end{array} \\ \begin{array}{rl}\int cosxlnsinxdx& =sinxlnsinx-\int cosxdx\\ & =sinxlnsinx-sinx+C\end{array} \end{gathered}$$

$\int e^x\ln (1+{\mathrm{e}}^x)dx$ (15)

$\begin{array}{rl}& u=\ln (1+e^x)dv=e^{x}dx\\ & du=\frac{e^x}{1+e^x}dxv=e^x\\ & \int e^x\ln (1+e^x)dx=e^x\ln (1+e^x)-\int \frac{e^{2x}}{1+e^x}dx\\ & =e^x\ln (1+e^x)-\int (e^x+\frac{-1}{1+e^x})dx\\ & =e^x\ln (1+e^x)-\int (e^x+\frac{-e^{-x}}{e^{-x}+1})dx\\ & =e^x\ln (1+e^x)-e^x-\ln (1+e^{-x})+C\end{array}$

أجد قيمة كل من التكاملات الآتية:

${\int }_0^{\pi /2}{e}^x\cos xdx$ (16)

$\begin{array}{rl}& \int e^x\cos xdx=\frac{1}{2}{e}^x(\sin x+\cos x)+C\\ & \Rightarrow {\int }_0^{\frac{\pi }{2}}{e}^x\cos xdx={\frac{1}{2}{e}^x(\sin x+\cos x)|}_0^{\frac{\pi }{2}}\\ & =\frac{1}{2}{e}^{\frac{\pi }{2}}-\frac{1}{2}{e}^0=\frac{1}{2}{e}^{\frac{\pi }{2}}-\frac{1}{2}\end{array}$

${\int }_1^e\ln x^{2}dx$ (17)

$\begin{array}{rl}& {\int }_1^e\ln x^{2}dx={\int }_1^{e}2\ln xdx\\ & u=2\ln xdv=dx\\ & du=\frac{2}{x}dxv=x\\ & {\int }_1^{e}2\ln xdx={2x\ln x|}_1^e-{\int }_1^{e}2dx\\ & ={2x\ln x|}_1^e-{2x|}_1^e\\ & =2e\ln e-2\ln 1-2e+2=2e-0-2e+2=2\end{array}$

${\int }_1^2\ln \left(x{e}^x\right)dx$ (18)

$\begin{array}{rl}{\int }_1^2\ln \left(x{e}^x\right)dx& ={\int }_1^2(\ln x+\ln e^x)dx\\ & ={\int }_1^2(\ln x+x)dx={\int }_1^2\ln xdx+{\int }_1^{2}xdx\end{array}$

نجد بطريقة ${\int }_1^2\ln xdx$ الأجزاء:

$$\begin{gathered} \begin{array}{ll}u=lnx& dv=dx\\ du=\frac{1}{x}dx& v=x\end{array} \\ \begin{array}{rl}{\int }_1^{2}lnxdx={xlnx|}_1^2-{\int }_1^{2}dx={xlnx|}_1^2-{x|}_1^2& =2ln2-ln1-2+1\\ & =2ln2-1\end{array} \\ \begin{array}{rl}& {\int }_1^{2}xdx={\frac{1}{2}{x}^2|}_1^2=\frac{4}{2}-\frac{1}{2}=\frac{3}{2}\\ & \Rightarrow {\int }_1^{2}ln\left(x{e}^x\right)dx=2ln2-1+\frac{3}{2}=2ln2+\frac{1}{2}\end{array} \end{gathered}$$

${\int }_{\pi /12}^{\pi /9}x{\sec }^{2}3xdx$ (19)

$$\begin{gathered} \begin{array}{ll}u=x& dv=se{c}^{2}3xdx\\ du=dx& v=\frac{1}{3}tan3x\end{array} \\ \begin{array}{rl}{\int }_{\frac{\pi }{12}}^{\frac{\pi }{9}}xse{c}^{2}3xdx& ={\frac{1}{3}xtan3x|}_{\frac{\pi }{12}}^{\frac{\pi }{9}}-{\int }_{\frac{\pi }{12}}^{\frac{\pi }{9}}\frac{1}{3}tan3xdx\\ & ={\frac{1}{3}xtan3x|}_{\frac{\pi }{12}}^{\frac{\pi }{9}}-{\int }_{\frac{\pi }{12}}^{\frac{\pi }{9}}\frac{1}{3}\frac{sin3x}{cos3x}dx\\ & ={\frac{1}{3}xtan3x|}_{\frac{\pi }{12}}^{\frac{\pi }{9}}+{\frac{1}{9}lncos3x|}_{\frac{\pi }{12}}^{\frac{\pi }{9}}\\ & =\frac{\pi }{27}tan\frac{\pi }{3}-\frac{\pi }{36}tan\frac{\pi }{4}+\frac{1}{9}lncos\frac{\pi }{3}-\frac{1}{9}lncos\frac{\pi }{4}\\ & =\frac{\pi \sqrt{3}}{27}-\frac{\pi }{36}+\frac{1}{9}ln\frac{1}{2}-\frac{1}{9}ln\frac{1}{\sqrt{2}}\end{array} \end{gathered}$$

${\int }_1^e{x}^4\ln xdx$ (20)

$\begin{array}{rl}& u=\ln xdv=x^{4}dx\\ & du=\frac{dx}{x}v=\frac{1}{5}{x}^5\\ & {\int }_1^e{x}^4\ln xdx={\frac{1}{5}{x}^5\ln x|}_1^e-{\int }_1^e\frac{1}{5}{x}^{4}dx\\ & ={\frac{1}{5}{x}^5\ln x|}_1^e-{\frac{1}{25}{x}^5|}_1^e\\ & =\frac{1}{5}{e}^5-0-\frac{1}{25}{e}^5+\frac{1}{25}=\frac{4e^5+1}{25}\end{array}$

${\int }_0^{\pi /2}{x}^2\sin xdx$ (21)

نجد $\int x^2\sin xdx$ باستخدام طريقة الجدول:

$\begin{array}{c}\int x^2\sin xdx=-x^2\cos x+2x\sin x+2\cos x+C\\ \Rightarrow {\int }_0^{\frac{\pi }{2}}{x}^2\sin xdx=-x^2\cos x+2x\sin x+{2\cos x|}_0^{\frac{\pi }{2}}\\ =\pi -2\end{array}$

${\int }_0^{1}x(e^{-2x}+e^{-x})dx$ (22)

$\begin{array}{rl}& u=xdv=(e^{-2x}+e^{-x})dx\\ & du=dxv=-\frac{1}{2}{e}^{-2x}-e^{-x}\\ & {\int }_0^{1}x(e^{-2x}+e^{-x})dx=-\frac{1}{2}x{e}^{-2x}-{x{e}^{-x}|}_0^1-{\int }_0^1(-\frac{1}{2}{e}^{-2x}-e^{-x})dx\\ & =-\frac{1}{2}x{e}^{-2x}-{x{e}^{-x}|}_0^1-{(\frac{1}{4}{e}^{-2x}+e^{-x})|}_0^1\\ & =-\frac{1}{2}{e}^{-2}-e^{-1}-\frac{1}{4}{e}^{-2}-e^{-1}+\frac{1}{4}+1\\ & =-\frac{3}{4}{e}^{-2}-2e^{-1}+\frac{5}{4}\end{array}$

${\int }_0^1\frac{x{e}^x}{(1+x{)}^2}dx$ (23)

$$\begin{gathered} \begin{array}{rl}& u=x{e}^{x}dv=(1+x{)}^{-2}dx\\ & du=(x{e}^x+e^x)dx=e^x(x+1)dxv=-(1+x{)}^{-1}\end{array} \\ \begin{array}{rl}{\int }_0^1\frac{x{e}^x}{(1+x{)}^2}dx& =-{x{e}^x(1+x{)}^{-1}|}_0^1+{\int }_0^1\frac{e^x(x+1)}{(1+x)}dx\\ & =-{\frac{x{e}^x}{1+x}|}_0^1+{e^x|}_0^1\\ & =-\frac{e}{2}+e-1=\frac{1}{2}e-1\end{array} \end{gathered}$$

${\int }_0^{1}x{3}^{x}dx$ (24)

$\begin{array}{rl}& u=xdv=3^{x}dx\\ & du=dxv=\frac{3^x}{\ln 3}\\ & {\int }_0^{1}x{3}^{x}dx={x\frac{3^x}{\ln 3}|}_0^1-{\int }_0^1\frac{3^x}{\ln 3}dx\\ & ={x\frac{3^x}{\ln 3}|}_0^1-{\frac{3^x}{(\ln 3{)}^2}|}_0^1\\ & =\frac{3}{\ln 3}-\frac{3}{(\ln 3{)}^2}+\frac{1}{(\ln 3{)}^2}=\frac{3\ln 3-2}{(\ln 3{)}^2}\end{array}$

أجد كلاً من التكاملات الآتية:

$\int x^3{e}^{x^2}dx$ (25)

$$\begin{gathered} \begin{array}{rl}& y=x^2\Rightarrow dx=\frac{dy}{2x}\\ & \int x^3{e}^{x^2}dx=\int x^3{e}^y\frac{dy}{2x}=\int \frac{1}{2}{x}^2{e}^{y}dy=\int \frac{1}{2}y{e}^{y}dy\\ & u=\frac{1}{2}ydv=e^{y}dy\\ & du=\frac{1}{2}dyv=e^y\end{array} \\ \begin{array}{rl}& \int \frac{1}{2}y{e}^{y}dy=\frac{1}{2}y{e}^y-\int \frac{1}{2}{e}^{y}dy\\ & =\frac{1}{2}y{e}^y-\frac{1}{2}{e}^y+C\\ & \int x^3{e}^{x^2}dx=\frac{1}{2}{x}^2{e}^{x^2}-\frac{1}{2}{e}^{x^2}+C\end{array} \end{gathered}$$

$\int \cos (\ln x)dx$ (26)

$$\begin{gathered} \begin{array}{rl}& y=lnx\Rightarrow \frac{dy}{dx}=\frac{1}{x}\Rightarrow dx=xdy,x=e^y\\ & \int cos(lnx)dx=\int xcosydy=\int e^{y}cosydy\end{array} \\ \begin{array}{rl}& \int e^{y}cosydy=\frac{1}{2}{e}^y(siny+cosy)+C\\ & \Rightarrow \int cos(lnx)dx=\frac{1}{2}{e}^{lnx}(sinlnx+coslnx)+C\\ & =\frac{1}{2}x(sinlnx+coslnx)+C\end{array} \end{gathered}$$

$\int x^3\sin x^{2}dx$ (27)

$$\begin{gathered} \begin{array}{rl}& y=x^2\Rightarrow dx=\frac{dy}{2x}\\ & \int x^{3}sin{x}^{2}dx=\int x^{3}siny\frac{dy}{2x}=\int \frac{1}{2}{x}^{2}sinydy=\int \frac{1}{2}ysinydy\\ & u=\frac{1}{2}ydv=sinydy\\ & du=\frac{1}{2}dyv=-cosy\end{array} \\ \begin{array}{rl}\int \frac{1}{2}ysinydy& =-\frac{1}{2}ycosy+\int \frac{1}{2}cosydy\\ & =-\frac{1}{2}ycosy+\frac{1}{2}siny+C\\ \int x^{3}sin{x}^{2}dx& =-\frac{1}{2}{x}^{2}cos{x}^2+\frac{1}{2}sin{x}^2+C\end{array} \end{gathered}$$

$\int e^{\cos x}\sin 2xdx$ (28)

$\begin{array}{rl}& y=\cos x\Rightarrow dx=\frac{ay}{-\sin x}\\ & \int e^{\cos x}\sin 2xdx=\int e^y(2\sin x\cos x)\frac{dy}{-\sin x}=\int -2y{e}^{y}dy\\ & u=-2ydv=e^{y}dy\\ & du=-2dyv=e^y\\ & \int -2y{e}^{y}dy=-2y{e}^y+\int 2e^{y}dy\\ & =-2y{e}^y+2e^y+C\\ & \Rightarrow \int e^{\cos x}\sin 2xdx=-2\cos x{e}^{\cos x}+2e^{\cos x}+C\end{array}$

$\int \sin \sqrt{x}dx$ (29)

$\begin{array}{rl}& y=\cos x\Rightarrow dx=\frac{ay}{-\sin x}\\ & \int e^{\cos x}\sin 2xdx=\int e^y(2\sin x\cos x)\frac{dy}{-\sin x}=\int -2y{e}^{y}dy\\ & u=-2ydv=e^{y}dy\\ & du=-2dyv=e^y\\ & \int -2y{e}^{y}dy=-2y{e}^y+\int 2e^{y}dy\\ & =-2y{e}^y+2e^y+C\\ & \Rightarrow \int e^{\cos x}\sin 2xdx=-2\cos x{e}^{\cos x}+2e^{\cos x}+C\end{array}$

$\int \frac{x^3{e}^{x^2}}{{(x^2+1)}^2}dx$ (30)

$$\begin{gathered} \begin{array}{rl}& y=x^2\Rightarrow \frac{dy}{dx}=2x\Rightarrow dx=\frac{dy}{2x}\\ & \int \frac{x^3{e}^{x^2}}{{(x^2+1)}^2}dx=\int \frac{x^3{e}^y}{(y+1{)}^2}\frac{dy}{2x}=\int \frac{1}{2}{x}^2\frac{e^y}{(y+1{)}^2}dy=\int \frac{\frac{1}{2}y{e}^y}{(y+1{)}^2}dy\end{array} \\ \begin{array}{rl}& u=\frac{1}{2}y{e}^{y}dv=\frac{1}{(y+1{)}^2}dy\\ & du=\frac{1}{2}(y{e}^y+e^y)dy=\frac{1}{2}{e}^y(y+1)dyv=\frac{-1}{y+1}\end{array} \\ \begin{array}{rl}\int \frac{\frac{1}{2}y{e}^y}{(y+1{)}^2}dy& =\frac{-y{e}^y}{2(y+1)}+\int \frac{1}{y+1}\times \frac{1}{2}{e}^y(y+1)dy\\ & =\frac{-y{e}^y}{2(y+1)}+\frac{1}{2}\int e^{y}dy\\ & =\frac{-y{e}^y}{2(y+1)}+\frac{1}{2}{e}^y+C\\ \int \frac{x^3{e}^{x^2}}{{(x^2+1)}^2}dx& =\frac{-x^2{e}^{x^2}}{2(x^2+1)}+\frac{1}{2}{e}^{x^2}+C=\frac{e^{x^2}}{2(x^2+1)}+C\end{array} \end{gathered}$$

إذا كان الشكل المجاور يمثل منحنى الاقتران: $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{x}$، حيث: $\mathit{x}\mathbf{\ge }\mathbf{0}$ فأجيب عن الأسئلة الثلاثة الآتية تباعاً:

(31) أجد إحداثيي كل من النقطة $A$، والنقطة $B$.

 الإحداثيان x للنقطتين A وB هما أصغر حلين موجبين للمعادلة:

$\begin{array}{rl}& f\left(x\right)=e^{-x}\sin 2x=0\\ & \Rightarrow \sin 2x=0\Rightarrow 2x=\pi ,2\pi ,\dots \\ & \Rightarrow x=\frac{\pi }{2},\pi ,\dots \\ & \Rightarrow A(\frac{\pi }{2},0),B(\pi ,0)\end{array}$

(32) أجد مساحة المنطقة المظللة.

$A={\int }_0^{\frac{\pi }{2}}{e}^{-x}\sin 2xdx+(-{\int }_{\frac{\pi }{2}}^{\pi }{e}^{-x}\sin 2xdx)$

للبسيط سنجد أولاً: $\int e^{-x}\sin 2xdx$ (التكامل غير المحدود)

$$\begin{gathered} \begin{array}{ll}u=e^{-x}& dv=sin2xdx\\ du=-e^{-x}dx& v=-\frac{1}{2}cos2x\end{array} \\ \int e^{-x}sin2xdx=-\frac{1}{2}{e}^{-x}cos2x-\int \frac{1}{2}{e}^{-x}cos2xdx \\ \begin{array}{rl}& u=\frac{1}{2}{e}^{-x}dv=cos2xdx\\ & du=-\frac{1}{2}{e}^{-x}dxv=\frac{1}{2}sin2x\end{array} \\ \begin{array}{rl}& \int e^{-x}sin2xdx=-\frac{1}{2}{e}^{-x}cos2x-\frac{1}{4}{e}^{-x}sin2x-\frac{1}{4}\int e^{-x}sin2xdx\\ & \int e^{-x}sin2xdx+\frac{1}{4}\int e^{-x}sin2xdx=-\frac{1}{2}{e}^{-x}cos2x-\frac{1}{4}{e}^{-x}sin2x\\ & \frac{5}{4}\int e^{-x}sin2xdx=-\frac{1}{2}{e}^{-x}cos2x-\frac{1}{4}{e}^{-x}sin2x+C\\ & \int e^{-x}sin2xdx=-\frac{1}{5}{e}^{-x}(2cos2x+sin2x)+C\\ & \Rightarrow A=-{\frac{1}{5}{e}^{-x}(2cos2x+sin2x)|}_0^{\frac{\pi }{2}}+{\frac{1}{5}{e}^{-x}(2cos2x+sin2x)|}_{\frac{\pi }{2}}^{\pi }\\ & =\frac{2}{5}{e}^{-\frac{\pi }{2}}+\frac{2}{5}+\frac{2}{5}{e}^{-\pi }+\frac{2}{5}{e}^{-\frac{\pi }{2}}\\ & =\frac{2}{5}(1+e^{-\pi }+2e^{-\frac{\pi }{2}})\end{array} \end{gathered}$$

(33) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $v\left(t\right)=t{e}^{-\mathrm{t}/2}$، حيث $t$ الزمن بالثواني، و$v$ سرعته المتجهة بالمتر لكل ثانية. إذا بدأ الجسيم الحركة من نقطة الأصل، فأجد موقعه بعد $t$ ثانية.

$\begin{array}{rl}& s\left(t\right)=\int t{e}^{-\frac{t}{2}}dt\\ & u=tdv=e^{-\frac{t}{2}}dt\\ & du=dtv=-2e^{-\frac{t}{2}}\\ & s\left(t\right)=-2t{e}^{-\frac{t}{2}}-\int -2e^{-\frac{t}{2}}dt=-2t{e}^{-\frac{t}{2}}-4e^{-\frac{t}{2}}+C\\ & s\left(0\right)=0-4+C\\ & 0=0-4+C\Rightarrow C=4\\ & \Rightarrow s\left(t\right)=-2t{e}^{-\frac{t}{2}}-4e^{-\frac{t}{2}}+4\end{array}$

في كل مما يأتي المشتقة الأولى للاقتران $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$، ونقطة يمر بها منحنى $\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$. أستعمل المعلومات المعطاة لإيجاد قاعدة الاقتران $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$:

$f^{\mathrm{\prime }}\left(x\right)=(x+2)\sin x;(0,2)$ (34)

$\begin{array}{rl}& f\left(x\right)=\int (x+2)\sin xdx\\ & u=x+2dv=\sin xdx\\ & du=dxv=-\cos x\\ & f\left(x\right)=-(x+2)\cos x+\int \cos xdx\\ & =-(x+2)\cos x+\sin x+C\\ & f\left(0\right)=-2+0+C\\ & 2=-2+0+C\Rightarrow C=4\\ & f\left(x\right)=-(x+2)\cos x+\sin x+4\end{array}$

$f^{\mathrm{\prime }}\left(x\right)=2x{e}^{-x};(0,3)$ (35)

$\begin{array}{rl}& f\left(x\right)=\int 2x{e}^{-x}dx\\ & u=2xdv=e^{-x}dx\\ & du=2dxv=-e^{-x}\\ & f\left(x\right)=-2x{e}^{-x}+\int 2e^{-x}dx\\ & =-2x{e}^{-x}-2e^{-x}+C\\ & f\left(0\right)=0-2+C\\ & 3=-2+C\Rightarrow C=5\\ & f\left(x\right)=-2x{e}^{-x}-2e^{-x}+5\end{array}$

(36) دورة تدريبية: تقدمت دعاء لدورة تدريبية متقدمة في الطباعة. إذا كان عدد الكلمات التي تطبعها دعاء في الدقيقة يزداد بمعدل: $N^{\mathrm{\prime }}\left(t\right)=(t+6)e^{-0.25t}$، حيث $N\left(t\right)$ عدد الكلمات التي تطبعها دعاء في الدقيقة بعد $t$ أسبوعاً من التحاقها بالدورة، فأجد $N\left(t\right)$، علماً بأن دعاء كانت تطبع 40 كلمة في الدقيقة عند بدء الدورة.

$\begin{array}{rl}& N\left(t\right)=\int (t+6)e^{-0.25t}dt\\ & u=t+6dv=e^{-0.25t}dt\\ & du=dtv=-4e^{-0.25t}\\ & N\left(t\right)=-4(t+6)e^{-0.25t}+\int 4e^{-0.25t}dt\\ & =-4(t+6)e^{-0.25t}-16e^{-0.25t}+C\\ & N\left(0\right)=-24-16+C\\ & 40=-24-16+C\Rightarrow C=80\\ & \Rightarrow N\left(t\right)=-4(t+6)e^{-0.25t}-16e^{-0.25t}+80\end{array}$