طباعة الدرس

أتدرب وأحل المسائل

التكامل المحدود

أجد قيمة كل من التكاملات الآتية:

(1) ${\int }_{-1}^{3}3x^2 dx$

$\begin{array}{rl}{\int }_{-1}^{3}3x^2 dx& ={x^3|}_{-1}^3\\ & =(3{)}^3-(-1{)}^3\\ & =28\end{array}$

(2) ${\int }_{-3}^{-2}6 dx$

$\begin{array}{rl}{\int }_{-3}^{-2}6 dx& ={6x|}_{-3}^{-2}\\ & =6(-2)-6(-3)\\ & =6\end{array}$

(3) ${\int }_0^2(3x^2+4x+3) dx$

$\begin{array}{rl}{\int }_0^2(3x^2+4x+3)& dx={(x^3+2x^2+3x)|}_0^2\\ =& \left(\right(2{)}^3+2(2{)}^2+3(2\left)\right)-\left(\right(0{)}^3+2(0{)}^2+3(0\left)\right)\\ =& 22\end{array}$

(4) ${\int }_1^{8}8\sqrt[3]{x} dx$

$\begin{array}{rl}{\int }_1^{8}8\sqrt[3]{x} dx& ={\int }_1^{8}8x^{\frac{1}{3}}dx\\ & ={6x^{\frac{4}{3}}|}_1^8\\ & ={6\sqrt[3]{x^4}|}_1^8\\ & =6\sqrt[3]{8^4}-6\sqrt[3]{0^4}\\ & =96\end{array}$

(5) ${\int }_1^9(\sqrt{x}-\frac{4}{\sqrt{x}}) dx$

$\begin{array}{rl}{\int }_1^9(\sqrt{x}-\frac{4}{\sqrt{x}}) dx& ={\int }_1^9(x^{\frac{1}{2}}-4x^{-\frac{1}{2}})dx\\ & ={(\frac{2}{3}{x}^{\frac{3}{2}}-8x^{\frac{1}{2}})|}_1^9\\ & ={(\frac{2}{3}\sqrt{x^3}-8\sqrt{x})|}_1^9\\ & =(\frac{2}{3}\sqrt{9^3}-8\sqrt{9})-(\frac{2}{3}\sqrt{1^3}-8\sqrt{1})\\ & =\frac{4}{3}\end{array}$

(6) ${\int }_{-2}^3(-x^2+4x-5) dx$

$\begin{array}{rl}& {\int }_{-2}^3(-x^2+4x-5) dx={(-\frac{1}{3}{x}^3+2x^2-5x)|}_{-2}^3\\ & =(-\frac{1}{3}(3{)}^3+2(3{)}^2-5(3\left)\right)-(-\frac{1}{3}(-2{)}^3+2(-2{)}^2-5(-2\left)\right)\\ & =-\frac{80}{3}\end{array}$

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

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

(8) ${\int }_{-3}^3(9-x^2) dx$

$\begin{array}{rl}{\int }_{-3}^3(9-x^2) dx& ={(9x-\frac{1}{3}{x}^3)|}_{-3}^3\\ & =\left(9\right(3)-\frac{1}{3}(3{)}^3)-\left(9\right(-3)-\frac{1}{3}(-3{)}^3)\\ & =36\end{array}$

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

$\begin{array}{rl}{\int }_1^4\frac{2+\sqrt{x}}{x^2} dx& ={\int }_1^4(\frac{2}{x^2}+\frac{\sqrt{x}}{x^2})dx\\ & ={\int }_1^4(2x^{-2}+x^{-\frac{3}{2}})dx\\ & ={(-2x^{-1}-2x^{-\frac{1}{2}})|}_1^4\\ & ={(\frac{-2}{x}-\frac{2}{\sqrt{x}})|}_1^4\\ & =(\frac{-2}{4}-\frac{2}{\sqrt{4}})-(\frac{-2}{1}-\frac{2}{\sqrt{1}})\\ & =\frac{5}{2}\end{array}$

(10) ${\int }_1^4{x}^3(\sqrt{x}+\frac{1}{x}) dx$

$\begin{array}{rl}{\int }_1^4{x}^3(\sqrt{x}+\frac{1}{x}) dx& ={\int }_1^4{x}^3(x^{\frac{1}{2}}+x^{-1})dx\\ & ={\int }_1^4(x^{\frac{7}{2}}+x^2)dx\\ & ={(\frac{2}{9}{x}^{\frac{9}{2}}+\frac{1}{3}{x}^3)|}_1^4\\ & ={(\frac{2}{9}\sqrt{x^9}+\frac{1}{3}{x}^3)|}_1^4\\ & =(\frac{2}{9}\sqrt{4^9}+\frac{1}{3}(4{)}^3)-(\frac{2}{9}\sqrt{1^9}+\frac{1}{3}(1{)}^3)\\ & =\frac{1211}{9}\end{array}$

(11) ${\int }_1^8(x^{1/3}-x^{-1/5}) dx$

$\begin{array}{rl}{\int }_1^8(x^{\frac{1}{3}}+x^{-\frac{1}{5}}) dx& ={(\frac{3}{4}{x}^{\frac{4}{3}}+\frac{5}{4}{x}^{\frac{4}{5}})|}_1^8\\ & ={(\frac{3}{4}\sqrt[3]{x^4}+\frac{5}{4}\sqrt[5]{x^4})|}_1^8\\ & =(\frac{3}{4}\sqrt[3]{8^4}+\frac{5}{4}\sqrt[5]{8^4})-(\frac{3}{4}\sqrt[3]{1^4}+\frac{5}{4}\sqrt[5]{1^4})\\ & =10+\frac{5}{4}\sqrt[5]{8^4}\end{array}$

(12) ${\int }_1^9(2+\sqrt{x}{)}^2 dx$

$\begin{array}{rl}{\int }_1^9(2+\sqrt{x}{)}^2 dx& ={\int }_1^9(4+4\sqrt{x}+x)dx\\ & ={\int }_1^9(4+4x^{\frac{1}{2}}+x)dx\\ & ={(4x+\frac{8}{3}{x}^{\frac{3}{2}}+\frac{1}{2}{x}^2)|}_1^9\\ & =\left(4\right(9)+\frac{8}{3}(9{)}^{\frac{3}{2}}+\frac{1}{2}\left(9{)}^2\right)-\left(4\right(1)+\frac{8}{3}(1{)}^{\frac{3}{2}}+\frac{1}{2}\left(1{)}^2\right)\\ & =\frac{424}{3}\end{array}$

(13) ${\int }_{-1}^4|3x-6| dx$

أعيد تعريف اقتران القيمة المطلقة:

$|3x-6|=\{\begin{array}{ll}6-3x,& x<2\\ 3x-6,& x\ge 2\end{array}$

بما أن الاقتران تشعب عند 2 ؛ فإنني أجزىء التكامل عنده:

$\begin{array}{rl}{\int }_{-1}^4|3x-6| dx& ={\int }_{-1}^2(6-3x)dx+{\int }_2^4(3x-6)dx\\ & ={(6x-\frac{3}{2}{x}^2)|}_{-1}^2+{(\frac{3}{2}{x}^2-6x)|}_2^4\\ =\left(6\right(2)-\frac{3}{2}(2{)}^2)& -\left(6\right(-1)-\frac{3}{2}(-1{)}^2)+\left(\frac{3}{2}\right(4{)}^2-6\left(4\right))-\left(\frac{3}{2}\right(2{)}^2-6\left(2\right))\\ & =\frac{39}{2}\end{array}$

(14) ${\int }_0^3|x-2| dx$

أعيد تعريف اقتران القيمة المطلقة:

$|x-2|=\{\begin{array}{ll}2-x,& x<2\\ x-2,& x\ge 2\end{array}$

بما أن الاقتران تشعب عند 2 ؛ فإنني أجزىء التكامل عنده:

$\begin{array}{rl}& {\int }_0^3|x-2|dx={\int }_0^2(2-x)dx+{\int }_2^3(x-2)dx\\ & ={(2x-\frac{1}{2}{x}^2)|}_0^2+{(\frac{1}{2}{x}^2-2x)|}_2^3\\ & =\left(2\right(2)-\frac{1}{2}(2{)}^2)-\left(2\right(0)-\frac{1}{2}(0{)}^2)+\left(\frac{1}{2}\right(3{)}^2-2\left(3\right))-\left(\frac{1}{2}\right(2{)}^2-2\left(2\right))\\ & =\frac{5}{2}\\ & \end{array}$

(15) ${\int }_2^3\frac{x^2-1}{x+1} dx$

$\begin{array}{rl}{\int }_2^3\frac{x^2-1}{x+1} dx& ={\int }_2^3\frac{(x+1)(x-1)}{x+1}dx\\ & ={\int }_2^3(x-1)dx\\ & ={(\frac{1}{2}{x}^2-x)|}_2^3\\ & =\left(\frac{1}{2}\right(3{)}^2-3)-\left(\frac{1}{2}\right(2{)}^2-2)=\frac{3}{2}\end{array}$