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إجابات كتاب التمارين

التكامل المحدود 

أجد قيمة كل من التكاملات الآتية:

${\int }_1^{5}10x^{-2}dx$ (1)

$\begin{array}{rl}{\int }_1^{5}10x^{-2}dx& =-{10x^{-1}|}_1^5\\ & =-{\frac{10}{x}|}_1^5=(-2)-(-10)=8\end{array}$

${\int }_0^2(2x^3-4x+5)dx$ (2)

$\begin{array}{rl}{\int }_0^2(2x^3-4x+5)dx& ={(\frac{1}{2}{x}^4-2x^2+5x)|}_0^2\\ & =8-8+10=10\end{array}$

${\int }_1^4\frac{x^3+2x^2}{\sqrt{x}}dx$ (3)

$\begin{array}{rl}{\int }_1^4\frac{x^3+2x^2}{\sqrt{x}}dx& ={\int }_1^4(\frac{x^3}{\sqrt{x}}+\frac{2x^2}{\sqrt{x}})dx\\ & ={\int }_1^4(x^{\frac{5}{2}}+2x^{\frac{3}{2}})dx\\ & ={(\frac{2}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}})|}_1^4\\ & =(\frac{256}{7}+\frac{128}{5})-(\frac{2}{7}+\frac{4}{5})=\frac{254}{7}+\frac{124}{5}=\frac{2138}{35}\end{array}$

${\int }_3^6{(x-\frac{3}{x})}^{2}dx$ (4)

$\begin{array}{rl}{\int }_3^6{(x-\frac{3}{x})}^{2}dx& ={\int }_3^6(x^2-6+\frac{9}{x^2})dx\\ & ={\int }_3^6(x^2-6+9x^{-2})dx\\ & ={(\frac{1}{3}{x}^3-6x-9x^{-1})|}_3^6\\ & ={(\frac{1}{3}{x}^3-6x-\frac{9}{x})|}_3^6\\ & =(72-36-\frac{3}{2})-(9-18-3)\\ & =\frac{93}{2}\end{array}$

${\int }_0^5\left(\right|x+3|-5)dx$ (5)

$\begin{array}{rl}& |x+3|=\{\begin{array}{c}-x-3,x<-3\\ x+3,x\ge -3\end{array}\\ & {\int }_0^5\left(\right|x+3|-5)dx={\int }_0^5(x+3-5)dx\\ & ={\int }_0^5(x-2)dx\\ & ={(\frac{1}{2}{x}^2-2x)|}_0^5=(\frac{25}{2}-10)-(0-0)=\frac{5}{2}\\ & \end{array}$

${\int }_0^{6}x(6-x)dx$ (6)

$\begin{array}{rl}{\int }_0^{6}x(6-x)dx& ={\int }_0^6(6x-x^2)dx\\ & ={(3x^2-\frac{1}{3}{x}^3)|}_0^6=(108-\frac{216}{3})-\left(0\right)=\frac{108}{3}=36\end{array}$

${\int }_1^2(6x-\frac{12}{x^4}+3)dx$ (7)

$\begin{array}{rl}{\int }_1^2(6x-\frac{12}{x^4}+3)dx& ={\int }_1^2(6x-12x^{-4}+3)dx\\ & ={(3x^2+4x^{-3}+3x)|}_1^2\\ & ={(3x^2+\frac{4}{x^3}+3x)|}_1^2\\ & =(12+\frac{1}{2}+6)-(3+4+3)=\frac{17}{2}\end{array}$

${\int }_0^7|2x-1|dx$ (8)

$\begin{array}{rl}& |2x-1|=\{\begin{array}{c}-2x+1,x<\frac{1}{2}\\ 2x-1,x\ge \frac{1}{2}\end{array}\\ & {\int }_0^7|2x-1|dx={\int }_0^{\frac{1}{2}}(-2x+1)dx+{\int }_{\frac{1}{2}}^7(2x-1)dx\\ & ={(-x^2+x)|}_0^{\frac{1}{2}}+{(x^2-x)|}_{\frac{1}{2}}^7\\ & =(-\frac{1}{4}+\frac{1}{2})-\left(0\right)+(49-7)-(\frac{1}{4}-\frac{1}{2})=\frac{85}{2}\end{array}$

${\int }_{-3}^4\left|x\right|dx$ (9)

$\begin{array}{rl}& \left|x\right|=\{\begin{array}{rl}-x& ,x<0\\ x,& x\ge 0\end{array}\\ & \begin{array}{rl}{\int }_{-3}^4\left|x\right|dx& ={\int }_{-3}^0-xdx+{\int }_0^{4}xdx\\ & =-{\frac{1}{2}{x}^2|}_{-3}^0+{\frac{1}{2}{x}^2|}_0^4\\ & =\left(0\right)-(-\frac{9}{2})+\left(8\right)-\left(0\right)=\frac{25}{2}\end{array}\end{array}$

${\int }_1^2\frac{x^2+x^3}{x}dx$ (10)

$\begin{array}{rl}{\int }_1^2\frac{x^2+x^3}{x}dx& ={\int }_1^2(\frac{x^2}{x}+\frac{x^3}{x})dx\\ & ={\int }_1^2(x+x^2)dx\\ & ={(\frac{1}{2}{x}^2+\frac{1}{3}{x}^3)|}_1^2=(2+\frac{8}{3})-(\frac{1}{2}+\frac{1}{3})=\frac{3}{2}+\frac{7}{3}=\frac{23}{6}\end{array}$

${\int }_3^4(6x^2-4x)dx$ (11)

$\begin{array}{rl}{\int }_3^4(6x^2-4x)dx& ={(2x^3+2x^2)|}_3^4\\ & =(128+32)-(54+18)=88\end{array}$

${\int }_{10}^{10}\frac{x+1}{x^2}dx$ (12)

${\int }_{10}^{10}\frac{x+1}{x^2}dx=0$

إذا كان $\mathbf{6}{\mathbf{\int }}_{\mathbf{-}\mathbf{3}}^{\mathbf{2}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{5}\mathbf{,}{\mathbf{\int }}_{\mathbf{-}\mathbf{3}}^{\mathbf{1}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{,}{\mathbf{\int }}_{\mathbf{-}\mathbf{3}}^{\mathbf{2}}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{2}$، فأجد كلّا مما يأتي:

${\int }_2^{2}f\left(x\right)dx$ (13)

${\int }_2^{2}f\left(x\right)dx=0$

${\int }_1^2\left(f\right(x)-5)dx$ (14)

$\begin{array}{rl}{\int }_1^2\left(f\right(x)-5)dx& ={\int }_1^{2}f\left(x\right)dx-{\int }_1^{2}5dx\\ & ={\int }_1^{-3}f\left(x\right)dx+{\int }_{-3}^{2}f\left(x\right)dx+{\int }_1^2-5dx\\ & =-4+5+{(-5x)|}_1^2\\ & =1+(-10)-(-5)\\ & =-4\end{array}$

${\int }_{-3}^2(-2f(x)+5g(x\left)\right)dx$ (15)

$\begin{array}{r}{\int }_{-3}^2(-2f(x)+5g(x\left)\right)dx=-2{\int }_{-3}^{2}f\left(x\right)dx+5{\int }_{-3}^{2}g\left(x\right)dx\\ =-2\left(5\right)+5(-2)=-20\end{array}$

${\int }_2^{-3}\left(g\right(x)+2x)dx$ (16)

$\begin{array}{rl}{\int }_2^{-3}\left(g\right(x)+2x)dx& ={\int }_2^{-3}g\left(x\right)dx+{\int }_2^{-3}2xdx\\ & =-(-2)+{\left(x^2\right)|}_2^{-3}=2+9-4=7\end{array}$

${\int }_2^{-3}\left(f\right(x)+g(x\left)\right)dx$ (17)

$\begin{array}{rl}{\int }_2^{-3}\left(f\right(x)+g(x\left)\right)dx& ={\int }_2^{-3}f\left(x\right)dx+{\int }_2^{-3}g\left(x\right)dx\\ & =-5+2=-3\end{array}$

${\int }_{-3}^2\left(4f\right(x)-3g(x\left)\right)dx$ (18)

$\begin{array}{rl}{\int }_{-3}^2\left(4f\right(x)-3g(x\left)\right)dx& =4{\int }_{-3}^{2}f\left(x\right)dx-3{\int }_{-3}^{2}g\left(x\right)dx\\ & =4\left(5\right)-3(-2)=26\end{array}$

(19) إذا كان $f\left(x\right)=\{\begin{array}{ll}{x}^2& ,x<2\\ 8-x& ,x\ge 2\end{array}$، فأجد قيمة${\int }_{-3}^{6}f\left(x\right)dx$.

$\begin{array}{rl}{\int }_{-3}^{6}f\left(x\right)dx& ={\int }_{-3}^{2}f\left(x\right)dx+{\int }_2^{6}f\left(x\right)dx\\ & ={\int }_{-3}^2{x}^{2}dx+{\int }_2^6(8-x)dx\\ & ={\left(\frac{1}{3}{x}^3\right)|}_{-3}^2+{(8x-\frac{1}{2}{x}^2)|}_2^6\\ & =\left(\frac{8}{3}\right)-(-9)+(48-18)-(16-2)=\frac{83}{3}\end{array}$

(20) سكان: أشارت دراسة إلى أن عدد السكان في إحدى القرى يتغير شهرياً بمعدل يمكن نمذجته بالاقتران: $P^{\mathrm{\prime }}\left(t\right)=5+3t^{2/3}$، حيث $t$ عدد الأشهر من الآن ، و$P\left(t\right)$ عدد السكان، أجد مقدار الزيادة في عدد سكان القرية في الأشهر الثمانية القادمة.

$\begin{array}{rl}P\left(t\right)& ={\int }_0^8(5+3t^{\frac{2}{3}})dt\\ & ={(5t+\frac{9}{5}{t}^{\frac{5}{3}})|}_0^8\\ & =(40+\frac{288}{5})-\left(0\right)\\ & =\frac{488}{5}\end{array}$

(21) إذا كان: ${\int }_2^3(x^2-a)dx=5$، فأجد قيمة الثابت $a$.

$\begin{array}{rl}& {\int }_2^3(x^2-a)dx=5\\ & {(\frac{1}{3}{x}^3-ax)|}_2^3=5\\ & (9-3a)-(\frac{8}{3}-2a)=5\\ & \frac{17}{3}-\alpha =5\\ & a=\frac{2}{3}\end{array}$