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التكامل

أجد كلاً من التكاملات الآتية:

$\int x^2{(2x^3+5)}^{4}dx$ (1)

$\begin{array}{rl}u=2x^3+5\Rightarrow & \frac{du}{dx}=6x^2\Rightarrow dx=\frac{du}{6x^2}\\ \int x^2{(2x^3+5)}^{4}dx& =\int x^2{u}^4\times \frac{du}{6x^2}\\ & =\int \frac{1}{6}{u}^{4}du\\ & =\frac{1}{30}{u}^5+C\\ & =\frac{1}{30}{(2x^3+5)}^5+C\end{array}$

$\int x^2\sqrt{x+3}dx$ (2)

$\begin{array}{rl}& u=x+3\Rightarrow dx=du,x=u-3\\ & \begin{array}{rl}{x}^2\sqrt{x+3}dx& =\int x^2\sqrt{u}du\\ & =\int (u-3{)}^2\sqrt{u}du\\ & =\int (u^{\frac{5}{2}}-6u^{\frac{3}{2}}+9u^{\frac{1}{2}})du\\ & =\frac{2}{7}{u}^{\frac{7}{2}}-\frac{12}{5}{u}^{\frac{5}{2}}+6u^{\frac{3}{2}}+C\\ & =\frac{2}{7}(x+3{)}^{\frac{7}{2}}-\frac{12}{5}(x+3{)}^{\frac{5}{2}}+6(x+3{)}^{\frac{3}{2}}+C\\ & =\frac{2}{7}\sqrt{(x+3{)}^7}-\frac{12}{5}\sqrt{(x+3{)}^5}+6\sqrt{(x+3{)}^3}+C\end{array}\end{array}$

$\int x(x+2{)}^{3}dx$ (3)

$\begin{array}{rl}& u=x+2\Rightarrow dx=du,x=u-2\\ & \int x(x+2{)}^{3}dx=\int x{u}^{3}du\\ & =\int (u-2)u^{3}du\\ & =\int (u^4-2u^3)du\\ & =\frac{1}{5}{u}^5-\frac{1}{2}{u}^4+C\\ & =\frac{1}{5}(x+2{)}^5-\frac{1}{2}(x+2{)}^4+C\\ & \end{array}$

$\int \frac{x}{\sqrt{x+4}}dx$ (4)

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

$\int \sin x\cos 2xdx$ (5)

$\begin{array}{rl}& \int \sin x\cos 2xdx=\int \sin x(2{\cos }^{2}x-1)dx\\ & u=\cos x\Rightarrow \frac{du}{dx}=-\sin x\Rightarrow dx=\frac{du}{-\sin x}\\ & \int \sin x\cos 2xdx=\int \sin x(2u^2-1)\times \frac{du}{-\sin x}\\ & =\int (1-2u^2)du\\ & =u-\frac{2}{3}{u}^3+C\\ & =\cos x-\frac{2}{3}{\cos }^{3}x+C\end{array}$

$\int \frac{e^{3x}}{e^x+1}dx$ (6)

$bbb\begin{array}{rl}u=e^x+1& \Rightarrow \frac{du}{dx}=e^x\Rightarrow dx=\frac{du}{e^x},e^x=u-1\\ \int \frac{e^{3x}}{e^x+1}dx& =\int \frac{e^{3x}}{u}\times \frac{du}{e^x}\\ & =\int \frac{e^{2x}}{u}du\\ & =\int \frac{(u-1{)}^2}{u}du\\ & =\int (u-2+\frac{1}{u})du\\ & =\frac{1}{2}{u}^2-2u+\ln \left|u\right|+C\\ & =\frac{1}{2}{(e^x+1)}^2-2(e^x+1)+\ln (e^x+1)+C\end{array}$

$\int {\sec }^{4}xdx$ (7)

$\begin{array}{rl}\int {\sec }^{4}xdx& =\int {\sec }^{2}x\times {\sec }^{2}xdx=\int {\sec }^{2}x(1+{\tan }^{2}x)dx\\ u=\tan x& \Rightarrow \frac{du}{dx}={\sec }^{2}x\Rightarrow dx=\frac{du}{{\sec }^{2}x}\\ \int {\sec }^{4}xdx& =\int {\sec }^{2}x(1+u^2)\times \frac{du}{{\sec }^{2}x}\\ & =\int (1+u^2)du\\ & =u+\frac{1}{3}{u}^3+C\\ & =\tan x+\frac{1}{3}{\tan }^{3}x+C\end{array}$

$\int \frac{\tan x}{{\cos }^{2}x}dx$ (8)

$\begin{array}{rl}\int \frac{\tan x}{{\cos }^{2}x}dx& =\int \tan x{\sec }^{2}xdx\\ u=\tan x& \Rightarrow \frac{du}{dx}={\sec }^{2}x\Rightarrow dx=\frac{du}{{\sec }^{2}x}\\ \int \frac{\tan x}{{\cos }^{2}x}dx& =\int u{\sec }^{2}x\times \frac{du}{{\sec }^{2}x}\\ & =\int udu\\ & =\frac{1}{2}{u}^2+C\\ & =\frac{1}{2}{\tan }^{2}x+C\end{array}$

$\int \frac{\sin (\ln x)}{x}dx$ (9)

$\begin{array}{rl}& u=\ln x\Rightarrow \frac{du}{dx}=\frac{1}{x}\Rightarrow dx=xdu\\ & \int \frac{\sin (\ln x)}{x}dx=\int \frac{\sin u}{x}\times xdu\\ & =\int \sin udu\\ & =-\cos u+C\\ & =-\cos (\ln x)+C\end{array}$

$\int \frac{\sin x\cos x}{1+{\sin }^{2}x}dx$ (10)

$\begin{array}{rl}\int \frac{\sin x\cos x}{1+{\sin }^{2}x}dx& =\frac{1}{2}\int \frac{2\sin x\cos x}{1+{\sin }^{2}x}dx\\ & =\frac{1}{2}\ln (1+{\sin }^{2}x)+C\end{array}$

$\int \frac{2e^x-2e^{-x}}{{(e^x+e^{-x})}^2}dx$ (11)

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

$\int \frac{-x}{(x+1)\sqrt{x+1}}dx$ (12)

$\begin{array}{rl}u=x+1⟹dx& =du,x=u-1\\ \int \frac{-x}{(x+1)\sqrt{x+1}}dx& =\int \frac{1-u}{u\sqrt{u}}du\\ & =\int \frac{1-u}{u^{\frac{3}{2}}}du\\ & =\int (u^{-\frac{3}{2}}-u^{-\frac{1}{2}})du\\ & =-2u^{-\frac{1}{2}}-2u^{\frac{1}{2}}+C\\ & =-2(x+1{)}^{-\frac{1}{2}}-2(x+1{)}^{\frac{1}{2}}+C\\ & =\frac{-2}{\sqrt{x+1}}-2\sqrt{x+1}+C\end{array}$

$\int x\sqrt[3]{x+10}dx$ (13)

$\begin{array}{rl}u=x+10\Rightarrow & dx=du,x=u-10\\ \int x\sqrt[3]{x+10}dx& =\int (u-10)u^{\frac{1}{3}}du\\ & =\int (u^{\frac{4}{3}}-10u^{\frac{1}{3}})du\\ & =\frac{3}{7}{u}^{\frac{7}{3}}-\frac{15}{2}{u}^{\frac{4}{3}}+C\\ & =\frac{3}{7}(x+10{)}^{\frac{7}{3}}-\frac{15}{2}(x+10{)}^{\frac{4}{3}}+C\\ & =\frac{3}{7}\sqrt[3]{(x+10{)}^7}-\frac{15}{2}\sqrt[3]{(x+10{)}^4}+C\end{array}$

$\int ({\sec }^2\frac{x}{2}{\tan }^7\frac{x}{2})dx$ (14)

$\begin{array}{rl}u=\tan \frac{x}{2}\Rightarrow \frac{du}{dx}& =\frac{1}{2}{\sec }^2\frac{x}{2}\Rightarrow dx=\frac{2du}{{\sec }^2\frac{x}{2}}\\ \int {\sec }^2\frac{x}{2}{\tan }^7\frac{x}{2}dx& =\int {\sec }^2\frac{x}{2}{u}^7\times \frac{2du}{{\sec }^2\frac{x}{2}}\\ & =2\int u^{7}du\\ & =\frac{1}{4}{u}^8+C\\ & =\frac{1}{4}{\tan }^8\frac{x}{2}+C\end{array}$

$\int \frac{{\sec }^{3}x+e^{\sin x}}{\sec x}dx$ (15)

$\begin{array}{rl}& \int \frac{{\sec }^{3}x+e^{\sin x}}{\sec x}dx=\int ({\sec }^{2}x+\cos x{e}^{\sin x})dx\\ & =\int {\sec }^{2}xdx+\int \cos x{e}^{\sin x}dx\\ & u=\sin x\Rightarrow \frac{du}{dx}=\cos x\Rightarrow dx=\frac{du}{\cos x}\\ & \int \frac{{\sec }^{3}x+e^{\sin x}}{\sec x}dx=\int {\sec }^{2}xdx+\int \cos x{e}^u\times \frac{du}{\cos x}\\ & =\tan x+\int e^{u}du\\ & =\tan x+e^u+C\\ & =\tan x+e^{\sin x}+C\end{array}$

$\int (1+\sqrt[3]{\sin x}){\cos }^{3}xdx$ (16)

$\begin{array}{rl}u=\sin x\Rightarrow \frac{du}{dx}=\cos x& \Rightarrow dx=\frac{du}{\cos x}\\ \int (1+\sqrt[3]{\sin x}){\cos }^{3}xdx& =\int (1+u^{\frac{1}{3}}){\cos }^{3}x\frac{du}{\cos x}\\ & =\int (1+u^{\frac{1}{3}}){\cos }^{2}xdu\\ & =\int (1+u^{\frac{1}{3}})(1-{\sin }^{2}x)du\\ & =\int (1+u^{\frac{1}{3}})(1-u^2)du\\ & =\int (1+u^{\frac{1}{3}})(1-u^2)du\\ & =\int (1-u^2+u^{\frac{1}{3}}-u^{\frac{7}{3}})du\\ & =u-\frac{1}{3}{u}^3+\frac{3}{4}{u}^{\frac{4}{3}}-\frac{3}{10}{u}^{\frac{10}{3}}+C\\ =\sin x-\frac{1}{3}{\sin }^{3}x+\frac{3}{4}& {\sin }^{\frac{4}{3}}x-\frac{3}{10}{\sin }^{\frac{10}{3}}x+C\end{array}$

$\int \sin x{\sec }^{5}xdx$ (17)

$\begin{array}{rl}& \int \sin x{\sec }^{5}xdx=\int \sin x{\cos }^{-5}xdx\\ & u=\cos x\Rightarrow \frac{du}{dx}=-\sin x\Rightarrow dx=\frac{du}{-\sin x}\\ & \int \sin x{\sec }^{5}xdx=\int \sin x{u}^{-5}\times \frac{du}{-\sin x}\\ & =-\int u^{-5}du\\ & =\frac{1}{4}{u}^{-4}+C\\ & =\frac{1}{4}{\cos }^{-4}x+C\\ & =\frac{1}{4}{\sec }^{4}x+C\end{array}$

$\int \frac{\sin x+\tan x}{{\cos }^{3}x}dx$ (18)

$\begin{array}{rl}\int \frac{\sin x+\tan x}{{\cos }^{3}x}dx& =\int (\tan x{\sec }^{2}x+\tan x{\sec }^{3}x)dx\\ & =\int \tan x\sec x(\sec x+{\sec }^{2}x)dx\\ u=\sec x\Rightarrow \frac{du}{dx}& =\tan x\sec x\Rightarrow dx=\frac{du}{\tan x\sec x}\\ \int \frac{\sin x+\tan x}{{\cos }^{3}x}dx& =\int \tan x\sec x(u+u^2)\frac{du}{\tan x\sec x}\\ & =\int (u+u^2)du\\ & =\frac{1}{2}{u}^2+\frac{1}{3}{u}^3+C\\ & =\frac{1}{2}{\sec }^{2}x+\frac{1}{3}{\sec }^{3}x+C\end{array}$

أجد قيمة كلا من التكاملات الآتية:

${\int }_0^{\pi /4}\sin x\sqrt{1-{\cos }^{2}2x}dx$ (19)

$\sqrt{1-{\cos }^{2}2x}=\sqrt{{\sin }^{2}2x}=|\sin 2x|$

لكن الزاوية $2x$ تكون ضمن الربع الأول عندما $0<x<\frac{\pi }{4}$

لذا فإن $\sin 2x>0$ ويكون $|\sin 2x|=\sin 2x$

$\begin{array}{rl}& {\int }_0^{\frac{\pi }{4}}\sin x\sqrt{1-{\cos }^{2}2x}dx={\int }_0^{\frac{\pi }{4}}\sin x\sin 2xdx\\ & ={\int }_0^{\frac{\pi }{4}}2{\sin }^{2}x\cos xdx\\ & u=\sin x\Rightarrow \frac{du}{dx}=\cos x\Rightarrow dx=\frac{du}{\cos x}\\ & x=0\Rightarrow u=0\\ & x=\frac{\pi }{4}\Rightarrow u=\frac{1}{\sqrt{2}}\\ & {\int }_0^{\frac{\pi }{4}}\sin x\sqrt{1-{\cos }^{2}2x}dx={\int }_0^{\frac{1}{\sqrt{2}}}2u^2\cos x\frac{du}{\cos x}\\ & ={\int }_0^{\frac{1}{\sqrt{2}}}2u^{2}du\\ & ={\frac{2}{3}{u}^3|}_0^{\frac{1}{\sqrt{2}}}\\ & =\frac{1}{3\sqrt{2}}\\ & \end{array}$

${\int }_0^{\pi /2}x\sin x^{2}dx$ (20)

$\begin{array}{rl}& u=x^2\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}\\ & x=\frac{\pi }{2}\Rightarrow u=\frac{{\pi }^2}{4}\\ & x=0\Rightarrow u=0\\ & {\int }_0^{\frac{\pi }{2}}x\sin x^{2}dx={\int }_0^{\frac{{\pi }^2}{4}}x\sin u\frac{du}{2x}\\ & =\frac{1}{2}{\int }_0^{\frac{{\pi }^2}{4}}\sin udu\\ & =-{\frac{1}{2}\cos u|}_0^{\frac{{\pi }^2}{4}}\\ & =-\frac{1}{2}(\cos \frac{{\pi }^2}{4}-1)\approx 0.891\end{array}$

${\int }_0^1\frac{x^3}{\sqrt{1+x^2}}dx$ (21)

$\begin{array}{rl}& u=1+x^2\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x},x^2=u-1\\ & x=0\Rightarrow u=1\\ & x=1\Rightarrow u=2\\ & {\int }_0^1\frac{x^3}{\sqrt{1+x^2}}dx={\int }_1^2\frac{x^3}{\sqrt{u}}\times \frac{du}{2x}=\frac{1}{2}{\int }_1^2\frac{x^2}{\sqrt{u}}du=\frac{1}{2}{\int }_1^2\frac{u-1}{\sqrt{u}}du\\ & =\frac{1}{2}{\int }_1^2(u^{\frac{1}{2}}-u^{-\frac{1}{2}})du\\ & ={\frac{1}{2}(\frac{2}{3}{u}^{\frac{3}{2}}-2u^{\frac{1}{2}})|}_1^2\\ & =\frac{1}{2}\left(\frac{2}{3}\right(2{)}^{\frac{3}{2}}-2(2{)}^{\frac{1}{2}}-\left(\frac{2}{3}\right(1)-2(1\left)\right))\\ & =\frac{2-\sqrt{2}}{3}\end{array}$

${\int }_0^{\pi /3}{\sec }^{2}x{\tan }^{5}xdx$ (22)

$\begin{array}{rl}& u=tanx\Rightarrow \frac{du}{dx}=se{c}^{2}x\Rightarrow dx=\frac{du}{se{c}^{2}x}\\ & x=0\Rightarrow u=0\\ & x=\frac{\pi }{3}\Rightarrow u=\sqrt{3}\\ & {\int }_0^{\frac{\pi }{3}}se{c}^{2}xta{n}^{5}xdx={\int }_0^{\sqrt{3}}se{c}^{2}x{u}^5\frac{du}{se{c}^{2}x}\\ & ={\int }_0^{\sqrt{3}}{u}^{5}du\\ & ={\frac{1}{6}{u}^6|}_0^{\sqrt{3}}\\ & =\frac{9}{2}\end{array}$

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

$\begin{array}{rl}& u=(x-1{)}^2\Rightarrow \frac{du}{dx}=2(x-1)\Rightarrow dx=\frac{du}{2(x-1)}\\ & x=0\Rightarrow u=1\\ & x=2\Rightarrow u=1\\ & {\int }_0^2(x-1)e^{(x-1{)}^2}dx={\int }_1^1(x-1)e^u\frac{du}{2(x-1)}=0\end{array}$

${\int }_1^4\frac{\sqrt{2+\sqrt{x}}}{\sqrt{x}}dx$ (24)

$\begin{array}{rl}& u=2+\sqrt{x}\Rightarrow \frac{du}{dx}=\frac{1}{2\sqrt{x}}\Rightarrow dx=2\sqrt{x}du\\ & x=1\Rightarrow u=3\\ & x=4\Rightarrow u=4\\ & {\int }_1^4\frac{\sqrt{2+\sqrt{x}}}{\sqrt{x}}dx={\int }_3^4\frac{\sqrt{u}}{\sqrt{x}}2\sqrt{x}du={\int }_3^{4}2\sqrt{u}du={\frac{4}{3}{u}^{\frac{3}{2}}|}_3^4=\frac{4(8-3\sqrt{3})}{3}\end{array}$

${\int }_0^1\frac{10\sqrt{x}}{{(1+\sqrt{x^3})}^2}dx$ (25)

$\begin{array}{rl}& u=1+x^{\frac{3}{2}}\Rightarrow \frac{du}{dx}=\frac{3}{2}{x}^{\frac{1}{2}}\Rightarrow dx=\frac{2}{3}\frac{du}{x^{\frac{1}{2}}}\\ & x=0\Rightarrow u=1\\ & x=1\Rightarrow u=2\\ & {\int }_0^1\frac{10\sqrt{x}}{{(1+\sqrt{x^3})}^2}dx={\int }_1^2\frac{10\sqrt{x}}{u^2}\frac{2}{3}\frac{du}{x^{\frac{1}{2}}}\\ & =\frac{20}{3}{\int }_1^2{u}^{-2}du=-{\frac{20}{3}{u}^{-1}|}_1^2=\frac{10}{3}\end{array}$

${\int }_0^{\pi /6}{2}^{\cos x}\sin xdx$ (26)

$\begin{array}{rl}& u=\cos x\Rightarrow \frac{du}{dx}=-\sin x\Rightarrow dx=\frac{du}{-\sin x}\\ & x=0\Rightarrow u=1\\ & x=\frac{\pi }{6}\Rightarrow u=\frac{\sqrt{3}}{2}\\ & {\int }_0^{\frac{\pi }{6}}{2}^{\cos x}\sin xdx={\int }_1^{\frac{\sqrt{3}}{2}}{2}^u\sin x\frac{du}{-\sin x}\\ & =-{\int }_1^{\frac{\sqrt{3}}{2}}{2}^{u}du\\ & =-{\frac{2^u}{\ln 2}|}_1^{\frac{\sqrt{3}}{2}}\\ & =-\frac{1}{\ln 2}(2^{\frac{\sqrt{3}}{2}}-2)\approx 0.256\\ & \end{array}$

${\int }_{\pi /4}^{\pi /2}{\csc }^{2}x{\cot }^{5}xdx$ (27)

$\begin{array}{rl}& u=\cot x\Rightarrow \frac{du}{dx}=-{\csc }^{2}x\Rightarrow dx=\frac{du}{-{\csc }^{2}x}\\ & x=\frac{\pi }{2}\Rightarrow u=0\\ & x=\frac{\pi }{4}\Rightarrow u=1\\ & {\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\csc }^{2}x{\cot }^{5}xdx={\int }_1^0{\csc }^{2}x{u}^5\frac{du}{-{\csc }^{2}x}\\ & ={\int }_1^0-u^{5}du\\ & =-{\frac{1}{6}{u}^6|}_1^0\\ & =\frac{1}{6}\end{array}$

أجد مساحة المنطقة المظللة في كل من التمثيلات البيانية الآتية: 

$\begin{array}{rl}& A=-{\int }_{-1}^{0}6x{(x^2+1)}^{3}dx+{\int }_0^{1}6x{(x^2+1)}^{3}dx\\ & u=x^2+1\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}\\ & x=-1\Rightarrow u=2\\ & x=0\Rightarrow u=1\\ & x=1\Rightarrow u=2\\ & A=-{\int }_2^{1}6x{u}^3\frac{du}{2x}+{\int }_1^{2}6x{u}^3\frac{du}{2x}\\ & ={\int }_1^{2}3u^{3}du+{\int }_1^{2}3u^{3}du={\int }_1^{2}6u^{3}du\\ & ={\frac{6}{4}{u}^4|}_1^2\\ & =\frac{45}{2}\end{array}$

$\begin{array}{rl}& A={\int }_2^4\frac{x}{(x-1{)}^3}dx\\ & u=x-1\Rightarrow dx=du,x=u+1\\ & x=2\Rightarrow u=1\\ & x=4\Rightarrow u=3\\ & \begin{array}{rl}A={\int }_2^4\frac{x}{(x-1{)}^3}dx& ={\int }_1^3\frac{u+1}{u^3}du\\ & ={\int }_1^3(u^{-2}+u^{-3})du={(-u^{-1}-\frac{1}{2}{u}^{-2})|}_1^3\\ & =-\frac{1}{3}-\frac{1}{2}\left(\frac{1}{9}\right)+1+\frac{1}{2}\\ & =\frac{10}{9}\end{array}\end{array}$

$\begin{array}{rl}& u=x^2⟹\frac{du}{dx}=2x⟹dx=\frac{du}{2x}\\ & x=-1\Rightarrow u=1\\ & x=0\Rightarrow u=0\\ & x=2\Rightarrow u=4\\ & A=-{\int }_{-1}^{0}x{e}^{x^2}dx+{\int }_0^{2}x{e}^{x^2}dx=-{\int }_1^{0}x{e}^u\frac{du}{2x}+{\int }_0^{4}x{e}^u\frac{du}{2x}\\ & =-{\int }_1^0\frac{1}{2}{e}^{u}du+{\int }_0^4\frac{1}{2}{e}^{u}du\\ & =-{\frac{1}{2}{e}^u|}_1^0+{\frac{1}{2}{e}^u|}_0^4\\ & =-\frac{1}{2}{e}^0+\frac{1}{2}{e}^1+\frac{1}{2}{e}^4-\frac{1}{2}{e}^0\\ & =\frac{1}{2}(e^4+e)-1\approx 27.658\end{array}$

$\begin{array}{rl}& u=x^2+\frac{\pi }{6}\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}\\ & x=\sqrt{\frac{\pi }{6}}\Rightarrow =\frac{\pi }{3}\\ & x=0\Rightarrow u=\frac{\pi }{6}\\ & \begin{array}{rl}A={\int }_0^{\sqrt{\frac{\pi }{6}}}2x& \cos (x^2+\frac{\pi }{6})dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}2x\cos u\frac{du}{2x}\\ & ={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\cos udu\\ & ={\sin u|}_{\frac{\pi }{6}}^{\frac{\pi }{3}}\\ & =\sin \frac{\pi }{3}-\sin \frac{\pi }{6}\\ & =\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}\approx 0.366\end{array}\end{array}$

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 $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$:

$f^{\mathrm{\prime }}\left(x\right)=2x{(4x^2-10)}^2;(2,10)$ (32)

$\begin{array}{rl}& f\left(x\right)=\int f^{\mathrm{\prime }}\left(x\right)dx=\int 2x{(4x^2-10)}^{2}dx\\ & u=4x^2-10\Rightarrow \frac{du}{dx}=8x\Rightarrow dx=\frac{du}{8x}\\ & f\left(x\right)=\int 2x{u}^2\frac{du}{8x}=\int u^2\frac{du}{4}\\ & =\frac{1}{4}\int u^{2}du\\ & =\frac{1}{12}{u}^3+C\\ & \Rightarrow f\left(x\right)=\frac{1}{12}{(4x^2-10)}^3+C\\ & f\left(2\right)=\frac{1}{12}\left(216\right)+C=10\Rightarrow C=-8\\ & 10=18+C\Rightarrow C=-8\\ & \Rightarrow f\left(x\right)=\frac{1}{12}{(4x^2-10)}^3-8\end{array}$

$f^{\mathrm{\prime }}\left(x\right)=x^2{e}^{-0.2x^3};(0,\frac{3}{2})$ (33)

$\begin{array}{rl}& f\left(x\right)=\int f^{\mathrm{\prime }}\left(x\right)dx=\int x^2{e}^{-0.2x^3}dx\\ & u=-0.2x^3\Rightarrow \frac{du}{dx}=-0.6x^2\Rightarrow dx=\frac{du}{-0.6x^2}\\ & f\left(x\right)=\int x^2{e}^u\frac{du}{-0.6x^2}=-\frac{10}{6}\int e^{u}du\\ & =-\frac{5}{3}{e}^u+C\\ & \Rightarrow f\left(x\right)=-\frac{5}{3}{e}^{-0.2x^3}+C\\ & f\left(0\right)=-\frac{5}{3}+C\\ & \frac{3}{2}=-\frac{5}{3}+C\Rightarrow C=\frac{19}{6}\\ & \Rightarrow f\left(x\right)=-\frac{5}{3}{e}^{-0.2x^3}+\frac{19}{6}\end{array}$

يبين الشكل المجاور جزءاً من منحنى الاقتران $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathit{x}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}{\mathbf{)}}^{\mathbf{4}}$: 

(34) أجد إحداثي نقطة تماس الاقتران مع المحور $x$

نجد أصفار الاقتران بحل المعادلة $f\left(x\right)=0$

$x(x-2{)}^4=0\Rightarrow x=0,x=2$

نقطة التقاطع $\left(0,0\right)$، فتكون نقطة التماس $(2,0)$

ويمكن التحقق بحساب $f^{\mathrm{\prime }}\left(2\right)$:

$\begin{array}{rl}& f^{\mathrm{\prime }}\left(x\right)=(x-2{)}^4+4x(x-2{)}^3\\ & f^{\mathrm{\prime }}\left(2\right)=(2-2{)}^4+4(2\left)\right(2-2{)}^3=0\end{array}$

(35) أجد مساحة المنطقة المحصورة بين منحنى الاقتران $f\left(x\right)$ والمحور $x$

$\begin{array}{rl}& A={\int }_0^{2}x(x-2{)}^{4}dx\\ & u=x-2\Rightarrow dx=du,x=u+2\\ & x=0\Rightarrow u=-2\\ & x=2\Rightarrow u=0\\ & \begin{array}{rl}A={\int }_0^{2}x(x-2{)}^{4}dx& ={\int }_{-2}^0(u+2)u^{4}du\\ & ={\int }_{-2}^0(u^5+2u^4)du={(\frac{1}{6}{u}^6+\frac{2}{5}{u}^5)|}_{-2}^0\\ & =0-\left(\frac{1}{6}\right(-2{)}^6+\frac{2}{5}(-2{)}^5)=\frac{32}{15}\end{array}\end{array}$

(36) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $v\left(t\right)=\sin \omega t{\cos }^2\omega t$، حيث $t$ الزمن بالثواني، و$v$ سرعته المتجهة بالمتر لكل ثانية، و$bbb$ ثابت، إذا انطلق الجسيم من نقطة الأصل، فأجد موقعه بعد $t$ ثانية. 

$\begin{array}{rl}& s\left(t\right)=\int \sin \omega t{\cos }^2\omega tdt\\ & u=\cos \omega t\Rightarrow \frac{du}{dx}=-\omega \sin \omega t\Rightarrow dt=\frac{du}{-\omega \sin \omega t}\\ & s\left(t\right)=\int \sin \omega t{u}^2\frac{du}{-\omega \sin \omega t}\\ & =\frac{-1}{\omega }\int u^{2}du=\frac{-1}{3\omega }{u}^3+C\\ & \Rightarrow s\left(t\right)=-\frac{1}{30}{\cos }^3\omega t+C\end{array}$

لكن $s\left(0\right)=0$ لأن الجسيم انطلق من نقطة الأصل.

$\begin{array}{rl}& s\left(0\right)=-\frac{1}{3\omega }+C\\ & 0=-\frac{1}{3\omega }+C\Rightarrow C=\frac{1}{3\omega }\\ & \Rightarrow s\left(t\right)=-\frac{1}{3\omega }{\cos }^3\omega t+\frac{1}{3\omega }\end{array}$

(37) طب: يمثل الاقتران $C\left(t\right)$ تركيز دواء في الدم بعد $t$ دقيقة من حقنه في جسم مريض، حيث $C$ مقيسة بالمليغرام لكل سنتيمتر مكعب $(\mathrm{mg}/{\mathrm{cm}}^3)$، إذا كان تركيز الدواء لحظة حقنه في جسم المريض $0.5\mathrm{mg}/{\mathrm{cm}}^3$، وأخذ يتغير بمعدل $C^{\mathrm{\prime }}\left(t\right)=\frac{-0.01e^{-0.01t}}{{(1+e^{-0.01t})}^2}$، فأجد $C\left(t\right)$. 

$\begin{array}{rl}& C\left(t\right)=\int C^{\mathrm{\prime }}\left(t\right)dt=\int \frac{-0.01e^{-0.01t}}{{(1+e^{-0.01t})}^2}dt\\ & u=1+e^{-0.01t}\Rightarrow \frac{du}{dt}=-0.01e^{-0.01t}\Rightarrow dt=\frac{du}{-0.01e^{-0.01t}}\\ & C\left(t\right)=\int \frac{-0.01e^{-0.01t}}{u^2}\times \frac{du}{-0.01e^{-0.01t}}=\int u^{-2}du\\ & =-u^{-1}+K\end{array}$

(استعمل الرمز $K$ لثابت التكامل بدل $C$ المعتاد لتمييز ثابت التكامل عن رمز الاقتران $C$):

$\begin{array}{rl}& C\left(t\right)=-{(1+e^{-0.01t})}^{-1}+K\\ & C\left(0\right)=-(2{)}^{-1}+K\\ & \frac{1}{2}=-\frac{1}{2}\Rightarrow K=1\\ & \Rightarrow C\left(t\right)=-{(1+e^{-0.01t})}^{-1}+1\\ & C\left(t\right)=\frac{-1}{1+e^{-0.01t}}+1\end{array}$

(38) أجد قيمة ${\int }_{\ln 3}^{\ln 4}\frac{e^{4x}}{e^x-2}dx$، ثم اكتب الإجابة بالصيغة الآتية: $\frac{a}{b}+c\ln d$، حيث $a,b,c,d$ ثوابت صحيحة.

$\begin{array}{rl}& u=e^x-2\Rightarrow \frac{du}{dx}=e^x\Rightarrow dx=\frac{du}{e^x}\\ & e^x=u+2\\ & x=\ln 3\Rightarrow u=e^{\ln 3}-2=3-2=1\\ & x=\ln 4\Rightarrow u=e^{\ln 4}-2=4-2=2\\ & \begin{array}{rl}{\int }_{\ln 3}^{\ln 4}\frac{e^{4x}}{e^x-2}dx& ={\int }_1^2\frac{e^{4x}}{u}\frac{du}{e^x}={\int }_1^2\frac{e^{3x}}{u}du\\ & ={\int }_1^2\frac{(u+2{)}^3}{u}du\\ & ={\int }_1^2\frac{u^3+6u^2+12u+8}{u}du\\ & ={\int }_1^2(u^2+6u+12+\frac{8}{u})du\end{array}\\ & ={(\frac{1}{3}{u}^3+3u^2+12u+8\ln |u\left|\right)|}_1^2\end{array}$

(39) إذا كان: $f^{\mathrm{\prime }}\left(x\right)=\tan x$، وكان: $f\left(3\right)=5$، فأثبت أن $f\left(x\right)=\ln \left|\frac{\cos 3}{\cos x}\right|+5$.

$\begin{array}{rl}& f\left(x\right)=\int \tan xdx=-\int \frac{-\sin x}{\cos x}dx=-\ln |\cos x|+C\\ & f\left(3\right)=-\ln |\cos 3|+C\\ & 5=-\ln |\cos 3|+C\Rightarrow C=5+\ln |\cos 3|\\ & f\left(x\right)=-\ln |\cos x|+5+\ln |\cos 3|=\ln \left|\frac{\cos 3}{\cos x}\right|+5\end{array}$