الأسئلة الوزارية حول التكامل

الأسئلة الوزارية حول التكامل

(1)- جد ناتج كل مما يأتي:

$\int (\sin x-3{\sec }^{2}x)dx$

$\int (\sin x-3{\sec }^{2}x)dx=-\cos x-3\tan x+c$

${\int }_0^3\frac{1}{\sqrt{1+x}}dx$

$\begin{array}{rl}{\int }_0^3\frac{1}{\sqrt{1+x}}dx& ={\int }_0^3(1+x{)}^{-\frac{1}{2}}dx={\left[\frac{(1+x{)}^{\frac{1}{2}}}{\frac{1}{2}}\right]}_0^3\\ & =[2\sqrt{1+x}{]}_0^3=[2\sqrt{4}]-[2\sqrt{1}]=4-2=2\end{array}$

$\int \cos 6x\cos 3xdx$

$\begin{array}{rl}& \int \cos 6x\cos 3xdx=\int (1-2{\sin }^{2}3x)\cos 3xdx\\ & =\int \cos 3xdx-2\int {\sin }^{2}3x\cos 3xdx\\ & =\frac{1}{3}\int \cos 3x\left(3\right)dx-\frac{2}{3}\int {\sin }^{2}3x\cos 3x\left(3\right)dx\\ & =\frac{1}{3}\sin 3x-\frac{2}{3}\cdot \frac{{\sin }^{3}3x}{3}+c=\frac{1}{3}\sin 3x-\frac{2}{9}{\sin }^{3}3x+c\end{array}$

$\int (\sec x-\sin x)(\sec x+\sin x)dx$

$\begin{array}{rl}& \int (\sec x-\sin x)(\sec x+\sin x)dx\\ & =\int ({\sec }^{2}x-{\sin }^{2}x)dx=\int {\sec }^{2}xdx-\frac{1}{2}\int (1-\cos 2x)dx\\ & =\tan x-\frac{1}{2}(x-\frac{1}{2}\sin 2x)+c=\tan x-\frac{1}{2}x+\frac{1}{4}\sin 2x+c\end{array}$

${\int }_4^{8}x\sqrt{x^2-15}dx={\int }_4^{8}x{(x^2-15)}^{\frac{1}{2}}dx$

$\begin{array}{rl}& {\int }_4^{8}x\sqrt{x^2-15}dx={\int }_4^{8}x{(x^2-15)}^{\frac{1}{2}}dx\\ & =\frac{1}{2}{\int }_4^{8}2x{(x^2-15)}^{\frac{1}{2}}dx={[\frac{1}{2}\cdot \frac{{(x^2-15)}^{\frac{3}{2}}}{\frac{3}{2}}]}_4^8={[\frac{1}{3}\cdot {(x^2-15)}^{\frac{3}{2}}]}_4^8\\ & =\left[\frac{1}{3}\right(64-15{)}^{\frac{3}{2}}]-\left[\frac{1}{3}\right(1{)}^{\frac{3}{2}}]=\left[\frac{1}{3}\right(49{)}^{\frac{3}{2}}]-\left[\frac{1}{3}\right(1{)}^{\frac{3}{2}}]=\left[\frac{1}{3}{\left(7^2\right)}^{\frac{3}{2}}\right]-\left[\frac{1}{3}\right(1{)}^{\frac{3}{2}}]\\ & =\frac{343}{3}-\frac{1}{3}=\frac{342}{3}=114\end{array}$

$\int \cos 2x{\sin }^{2}xdx$

$$\begin{gathered} \int \cos 2x{\sin }^{2}xdx=\int \cos 2x\cdot \frac{1}{2}(1-\cos 2x)dx \\ \begin{array}{rl}& =\frac{1}{2}\int \cos 2xdx-\frac{1}{2}\int {\cos }^{2}2xdx=\frac{1}{2}\cdot \frac{1}{2}\int \cos 2x\left(2\right)dx-\frac{1}{2}\cdot \frac{1}{2}\int (1+\cos 4x)dx\\ & =\frac{1}{4}\int \cos 2x\left(2\right)dx-\frac{1}{4}[\int dx+\frac{1}{4}\int \cos 4x(4)dx\\ & =\frac{1}{4}\sin 2x-\frac{1}{4}(x+\frac{1}{4}\sin 4x)+c=\frac{1}{4}\sin 2x-\frac{1}{4}x-\frac{1}{16}\sin 4x+c\end{array} \end{gathered}$$

$\int (1+\cos 3x{)}^{2}dx$

$\begin{array}{rl}& \int (1+\cos 3x{)}^{2}dx=\int (1+2\cos 3x+{\cos }^{2}3x)dx\\ & =\int dx+2\int \cos 3xdx+\frac{1}{2}\int (1+\sin 6x)dx\\ & =\int dx+2\int \cos 3xdx+\frac{1}{2}[\int dx+\frac{1}{6}\int \sin 6x(6\left)dx\right]\\ & =x+\frac{2}{3}\sin 3x+\frac{1}{2}(x+\frac{1}{6}\sin 6x)+c=\frac{3}{2}x+\frac{2}{3}\sin 3x+\frac{1}{12}\sin 6x+c\end{array}$

$\int (\cos x-\sin 2x{)}^{2}dx$

$\begin{array}{rl}& \int (\cos x-\sin 2x{)}^{2}dx=\int ({\cos }^{2}x-2\cos x\sin 2x+{\sin }^{2}2x)\\ & =\frac{1}{2}\int (1+\cos 2x)dx-2\int \cos x\cdot (2\sin x\cos x)dx+\frac{1}{2}\int (1-\cos 4x)dx\\ & =\frac{1}{2}[\int dx+\frac{1}{2}\int \cos 2x(2\left)dx\right]+4\int {\cos }^{2}x(-\sin x)dx+\frac{1}{2}[\int dx-\frac{1}{4}\int \cos 4x(4\left)dx\right]\\ & =\frac{1}{2}(x+\frac{1}{2}\sin 2x)+4\cdot \frac{{\cos }^{3}x}{3}+\frac{1}{2}(x-\frac{1}{4}\sin 4x)+c\\ & =\frac{1}{2}x+\frac{1}{4}\sin 2x+\frac{4}{3}{\cos }^{3}x+\frac{1}{2}x-\frac{1}{8}\sin 4x+c\\ & =x+\frac{1}{4}\sin 2x+\frac{4}{3}{\cos }^{3}x-\frac{1}{8}\sin 4x+c\end{array}$

(2)- إذا كان ${\int }_{-1}^a(x-x^3)dx=\frac{-9}{4}$ ماقيمة $a\in R^{+}$؟

$\begin{array}{rl}& {\int }_{-1}^a(x-x^3)dx=\frac{-9}{4}\Rightarrow {[\frac{x^2}{2}-\frac{x^4}{4}]}_{-1}^a=\frac{-9}{4}\Rightarrow [\frac{a^2}{2}-\frac{a^4}{4}]-[\frac{1}{2}-\frac{1}{4}]=\frac{-9}{4}\\ & \frac{a^2}{2}-\frac{a^4}{4}-\frac{1}{2}+\frac{1}{4}=\frac{-9}{4}\Rightarrow \frac{a^2}{2}-\frac{a^4}{4}-\frac{1}{4}=\frac{-9}{4}\stackrel{(x-4)}{⟹}-2a^2+a^4+1=9\\ & a^4-2a^2+1-9=0\Rightarrow a^4-2a^2-8=0\\ & (a^2-4)(a^2+2)=0\Rightarrow (a^2-4)=0\\ & \text{either }{a}^2=4\Rightarrow a=2\\ & \text{or}{a}^2+2=0 \left(\mathrm{تهمل}\right)\end{array}$

(3)- إذا كان ${\int }_a^b(2x+3)dx=12$ وكان $a+2b=3$ ما قيمة $a,b\in R$؟

$\begin{array}{rl}& a=3-2b\dots \dots \dots \left(1\right)\\ & {\int }_a^b(2x+3)dx=12\Rightarrow {[x^2+3x]}_a^b=12\Rightarrow [b^2+3b]-[a^2+3a]=12\\ & b^2+3b-(3-2b{)}^2-3(3-2b)=12\\ & b^2+3b-(9-12b+4b^2)-9+6b-12=0\\ & b^2+3b-9+12b-4b^2-9+6b-12=0\\ & [-3b^2+21b-30=0]÷-3\Rightarrow b^2-7b+10=0\\ & (b-5)(b-2)=0\\ & \text{either }b-2=0\Rightarrow b=2\Rightarrow a=3-2\left(2\right)\Rightarrow a=-1\\ & \text{or }b-5=0\Rightarrow b=5\Rightarrow a=3-2\left(5\right)\Rightarrow a=-7\end{array}$

(4)- جد المساحة المحددة بمنحني الدالة $f\left(x\right)=1-2{\sin }^{2}x$ ومحور السينات وعلى $[0,\frac{\pi }{2}]$.

$$\begin{gathered} \begin{array}{rl}& 1-2{\sin }^{2}x=0\Rightarrow \cos 2x=0\\ & 2x=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{4}\in [0,\frac{\pi }{2}]\\ & 2x=\frac{3\pi }{2}\Rightarrow x=\frac{3\pi }{4}\notin [0,\frac{\pi }{2}][0,\frac{\pi }{4}],[\frac{\pi }{4},\frac{\pi }{2}]=0 \left(\mathrm{الفترات}\right)\end{array} \\ \begin{array}{rl}A& =|{\int }_0^{\frac{\pi }{4}}\cos 2xdx|+|{\int }_{\frac{\pi }{4}}^{\frac{2}{2}}\cos 2xdx|=\left|{[\frac{1}{2}\sin 2x]}_0^{\frac{\pi }{4}}\right|+\left|{[\frac{1}{2}\sin 2x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}\right|\\ & =|[\frac{1}{2}\sin \frac{\pi }{2}]-[\frac{1}{2}\sin 0]|+|[\frac{1}{2}\sin \pi ]-[\frac{1}{2}\sin \frac{\pi }{2}]|\\ & =\left|\frac{1}{2}\right(1)-\frac{1}{2}(0\left)\right|+\left|\frac{1}{2}\right(0)-\frac{1}{2}(1\left)\right|=\frac{1}{2}+\frac{1}{2}=1{\text{unit}}^2\end{array} \end{gathered}$$

(5)- جد ناتج ما يأتي:

$\int {\sin }^{2}x{\cos }^{2}xdx$

$\begin{array}{rl}& \int {\sin }^{2}x{\cos }^{2}xdx=\int [\sin x\cos x{]}^{2}dx=\int {\left[\frac{1}{2}\right(2\sin x\cos x\left)\right]}^{2}dx\\ & =\int {[\frac{1}{2}\sin 2x]}^{2}dx=\int \frac{1}{4}{\sin }^{2}2xdx=\frac{1}{4}\cdot \frac{1}{2}\int (1-\cos 4x)dx\\ & =\frac{1}{8}[\int dx-\frac{1}{4}\int \cos 4x(4\left)dx\right]=\frac{1}{8}(x-\frac{1}{4}\sin 4x)+c=\frac{1}{8}x-\frac{1}{32}\sin 4x+c\end{array}$

(6)- جد ناتج ما يأتي:

${\int }_{-1}^1\frac{dx}{9-12x+4x^2}$

$$\begin{gathered} {\int }_{-1}^1\frac{dx}{9-12x+4x^2}={\int }_{-1}^1\frac{dx}{(3-2x{)}^2}={\int }_{-1}^1(3-2x{)}^{-2}dx \\ \begin{array}{rl}& =\frac{-1}{2}{\int }_{-1}^1(3-2x{)}^{-2}\cdot (-2)dx\\ & ={[\frac{-1}{2}\cdot \frac{(3-2x{)}^{-1}}{-1}]}_{-1}^1={\left[\frac{1}{2(3-2x)}\right]}_{-1}^1=\left[\frac{1}{2\left(1\right)}\right]-\left[\frac{1}{2\left(5\right)}\right]=\frac{1}{2}-\frac{1}{10}=\frac{5-1}{10}=\frac{2}{5}\end{array} \end{gathered}$$

(7)- جد المساحة المحددة بمنحني الدالة $y=x^3-9x$ ومحور السينات والفترة $[-3,3]$.

$$\begin{gathered} x^3-9x=0\Rightarrow x(x^2-9)=0 \\ \text{either }x=0 \\ \text{or }{x}^2-9=0\Rightarrow x^2=9\Rightarrow x=\pm 3\in [-3,3] \\ \begin{array}{rl}& A=\left|{\int }_{-3}^0(x^3-9x)dx\right|+\left|{\int }_0^3(x^3-9x)dx\right|=\left|{[\frac{x^4}{4}-\frac{9x^2}{2}]}_{-3}^0\right|+\left|{[\frac{x^4}{4}-\frac{9x^2}{2}]}_0^3\right|\\ & =\left|\right[0]-[\frac{81}{4}-\frac{81}{2}]|+|[\frac{81}{4}-\frac{81}{2}]-[0\left]\right|\\ & =|\frac{-81}{4}+\frac{81}{2}|+|\frac{81}{4}-\frac{81}{2}|=\left|\frac{-81+162}{4}\right|+\left|\frac{81-162}{4}\right|=\frac{81}{4}+\frac{81}{4}=\frac{162}{4}=40\frac{1}{2}\text{unit}\end{array} \end{gathered}$$

(8)- جد قيمة ما يأتي: ${\int }_0^4\sqrt{x^2+5x}(2x+5)dx$

$\begin{array}{rl}& ={\int }_0^4{(x^2+5x)}^{\frac{1}{2}}(2x+5)dx={\left[\frac{{(x^2+5x)}^{\frac{1}{2}}}{\frac{3}{2}}\right]}_0^4={\left[\frac{2}{3}{(x^2+5x)}^{\frac{3}{2}}\right]}_0^4\\ & =\left[\frac{2}{3}\right(16+20{)}^{\frac{3}{2}}]-\left[\frac{2}{3}\right(0{)}^{\frac{3}{2}}]=\left[\frac{2}{3}{\left(6^2\right)}^{\frac{3}{2}}\right]=\frac{2}{3}\left(216\right)=\frac{432}{3}=144\end{array}$

(9)- جد المساحة المحددة بمنحني الدالتين $y=3x^2,y=x^4-4$

$$\begin{gathered} \begin{array}{rl}& x^4-4=3x^2\Rightarrow h\left(x\right)=x^4-3x^2-4\Rightarrow x^4-3x^2-4=0\\ & (x^2-4)(x^2+1)=0\\ & \mathrm{either}(x^2-4)=0\Rightarrow x^2=4\Rightarrow x=\pm 2\\ & \text{or}(x^2+1)=0 \left(\mathrm{تهمل}\right)\end{array} \\ \begin{array}{rl}& A=\left|{\int }_{-2}^2(x^4-3x^2-4)dx\right|=\left|{[\frac{x^5}{5}-x^3-4x]}_{-2}^2\right|\\ & =|[\frac{32}{5}-8-8]-[\frac{-32}{5}+8+8]|=|\frac{64}{5}-32|=\frac{96}{5}{\text{unit}}^2\end{array} \end{gathered}$$

(10)- جد المساحة المحددة بمنحني الدالتين $y=2x,y=x^2$ وعلى الفترة $[1,3]$.

$$\begin{gathered} h\left(x\right)=x^2-2x\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0 \\ \text{either }x=0\notin [1,3] \\ \text{or }x-2=0⟹x=2\in [1,3] \\ \begin{array}{rl}& \therefore A=\left|{\int }_1^2(x^2-2x)dx\right|+\left|{\int }_2^3(x^2-2x)dx\right|=|{[\frac{x^3}{3}-x^2]}_1^2+{[\frac{x^3}{3}-x^2]}_2^3|\\ & =\left|\right|\frac{8}{3}-4]-[\frac{1}{3}-1]|+|[9-9]-[\frac{8}{3}-4]\mid \\ & =|\frac{8}{3}-4-\frac{1}{3}+1|+|-\frac{8}{3}+4|=|\frac{7}{3}-3|+\left|\frac{-8+12}{3}\right|=\left|\frac{7-9}{3}\right|+\left|\frac{4}{3}\right|=\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2\text{unit}\end{array} \end{gathered}$$

(11)- إذا كان ${\int }_h^4\frac{x}{\sqrt{x^2+9}}dx=2$ جد قيمة $h$.

$\begin{array}{rl}& {\int }_h^4\frac{x}{\sqrt{x^2+9}}dx=2\Rightarrow {\int }_h^{4}x{(x^2+9)}^{\frac{-1}{2}}dx=2\\ & \frac{1}{2}{\int }_h^{4}2x{(x^2+9)}^{\frac{-1}{2}}dx=2\Rightarrow {\left[\frac{1}{2}\frac{{(x^2+9)}^{\frac{1}{2}}}{\frac{1}{2}}\right]}_h^4=2\Rightarrow {\left[\sqrt{x^2+9}\right]}_h^4=2\\ & \left[\sqrt{16+9}\right]-\left[\sqrt{h^2+9}\right]=2\Rightarrow \sqrt{25}-\sqrt{h^2+9}=2\\ & \sqrt{h^2+9}=5-2\Rightarrow \sqrt{h^2+9}=3 \left(\mathrm{بالتربيع}\right)\\ & h^2+9=9\Rightarrow h^2=0\stackrel{\left(\mathrm{بالجذر}\right)}{⟹}h=0\end{array}$

(12)- جد قيمة ${\int }_1^2\frac{dx}{(5-2x{)}^2}$

$\begin{array}{rl}& {\int }_1^2(5-2x{)}^{-2}dx=\frac{-1}{2}{\int }_1^2(5-2x{)}^{-2}(-2)dx={[\frac{-1}{2}\cdot \frac{(5-2x{)}^{-1}}{-1}]}_1^2\\ & ={\left[\frac{1}{2(5-2x)}\right]}_1^2=\left[\frac{1}{2(5-4)}\right]-\left[\frac{1}{2(5-2)}\right]=\frac{1}{2}-\frac{1}{6}=\frac{3-1}{6}=\frac{2}{6}=\frac{1}{3}\end{array}$

(13)- جد قيمة ${\int }_1^2\frac{dx}{(3x-4{)}^2}$

$\begin{array}{rl}& ={\int }_1^2(3x-4{)}^{-2}dx=\frac{1}{3}{\int }_1^2(3x-4{)}^{-2}\left(3\right)dx={[\frac{1}{3}\cdot \frac{(3x-4{)}^{-1}}{-1}]}_1^2\\ & ={\left[\frac{-1}{3(3x-4)}\right]}_1^2=\left[\frac{-1}{3(6-4)}\right]-\left[\frac{-1}{3(3-4)}\right]=\frac{-1}{6}-\frac{1}{3}=\frac{-1-2}{6}=\frac{-3}{6}=\frac{-1}{2}\end{array}$

(14)- إذا كان ${\int }_c^{b}f\left(x\right)dx=3 ,\hspace{0.17em}{\int }_a^{b}f\left(x\right)dx=5$ وكانت $c\in [a,b]$، جد قيمة ${\int }_a^{c}f\left(x\right)dx$.

$\begin{array}{rl}& {\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx\Rightarrow 5={\int }_a^{c}f\left(x\right)dx+3\\ & \therefore {\int }_a^{c}f\left(x\right)dx=5-3\Rightarrow {\int }_a^{c}f\left(x\right)dx=2\end{array}$

(15)- جد $\int {\cos }^{2}2x\sin xdx$

$\begin{array}{rl}& =\int {(2{\cos }^{2}x-1)}^2\sin xdx=\int (4{\cos }^{4}x-4{\cos }^{2}x+1)\sin xdx\\ & =-4\int {\cos }^{4}x(-\sin x)dx+4\int {\cos }^{2}x\cdot (-\sin x)dx+\int \sin xdx\\ & =-4\frac{{\cos }^{5}x}{5}+4\cdot \frac{{\cos }^{3}x}{3}-\cos x+c=\frac{-4}{5}{\cos }^{5}x+\frac{4}{3}{\cos }^{3}x-\cos x+c\end{array}$

(16)- جسم يتحرك بسرعة $v\left(t\right)=3t^2-12t+9$ في أي زمن $t$ احسب:

المسافة المقطوعة خلال الفترة $[0,2]$.

$\begin{array}{rl}& [3t^2-12t+9=0]÷3\Rightarrow t^2-4t+3=0\\ & (t-3)(t-1)=0\\ & \text{either}t-3=0\Rightarrow t=3\notin [0,2]\\ & \text{or}t-1=0\Rightarrow t=1\in [0,2]\\ & \therefore d=\left|{\int }_0^1(3t^2-12t+9)dt\right|+\left|{\int }_1^2(3t^2-12t+9)dt\right|\\ & =\left|{[t^3-6t^2+9t]}_0^1\right|+\left|{[t^3-6t^2+9t]}_1^2\right|\\ & =\left|\right[1-6+9]-[0\left]\right|+\left|\right[8-24+18]-[1-6+9\left]\right|\\ & =\left|4\right|+|2-4|=4+2=6m\end{array}$

الزمن الذي يصبح فيه التعجيل $18\mathrm{m}\mathrm{\setminus }{min}^2$.

$\begin{array}{rl}& a\left(t\right)=V^{\mathrm{\prime }}\left(t\right)=6t-12\\ & 18=6t-12\Rightarrow 6t=18+12\Rightarrow 6t=30\Rightarrow t=5\text{min}\end{array}$

(17)- جد قيمة ${\int }_3^8\frac{x}{\sqrt{x^3+x^2}}dx$

$\begin{array}{rl}& ={\int }_3^8\frac{x}{\sqrt{x^2(x+1)}}dx={\int }_3^8\frac{x}{x(x+1{)}^{\frac{1}{2}}}dx={\int }_3^8(x+1{)}^{\frac{-1}{2}}dx\\ & ={\left[\frac{(x+1{)}^{\frac{1}{2}}}{\frac{1}{2}}\right]}_3^8=[2\sqrt{x+1}{]}_3^8=[2\sqrt{8+1}]-[2\sqrt{3+1}]\\ & =\left[2\right(3\left)\right]-\left[2\right(2\left)\right]=6-4=2\end{array}$

(18)- جد المساحة المحددة بين المنحيين: $y={\sin }^{2}x,y={\cos }^{2}x$ الفترة $[0,\frac{\pi }{2}]$.

$\begin{array}{rl}& h\left(x\right)={\cos }^{2}x-{\sin }^{2}x\Rightarrow h\left(x\right)=\cos 2x\\ & \cos 2x=0\Rightarrow 2x=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{4}\in [0,\frac{\pi }{2}]\\ & 2x=\frac{3\pi }{2}\Rightarrow \frac{3\pi }{4}\notin [0,\frac{\pi }{2}]\\ & \therefore A=|{\int }_0^{\frac{\pi }{4}}\cos 2xdx|+|{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cos 2xdx|=\left|{[\frac{1}{2}\sin 2x]}_0^{\frac{\pi }{4}}\right|+\left|{[\frac{1}{2}\sin 2x]}_{\frac{\pi }{4}}^{\frac{\pi }{2}}\right|\\ & =|[\frac{1}{2}\sin \frac{\pi }{2}]-[\frac{1}{2}\sin 0]|+|[\frac{1}{2}\sin \pi ]-[\frac{1}{2}\sin \frac{\pi }{2}]|\\ & =\left|\frac{1}{2}\right(1)-\frac{1}{2}(0\left)\right|+\left|\frac{1}{2}\right(0)-\frac{1}{2}(1\left)\right|=\frac{1}{2}+\frac{1}{2}=1 \left(\mathrm{مربعة} \mathrm{وحدة}\right)\end{array}$

(19)- جسم يتحرك على خط مستقيم بسرعة $v\left(t\right)=3t^2+4t+7\mathrm{m}/\mathrm{s}$ جد المسافة التي يقطعها الجسم بعد مضي $\text{(4)}$ ثواني من بدء الحركة ثم جد التعجيل عندها.

المسافة المقطوعة بعد مرور $t$ ثانية من بدء الحركة تكون الفترة $[0,t]$

$\begin{array}{rl}& v\left(t\right)=3t^2+4t+7\Rightarrow 3t^2+4t+7\ne 0\\ & d={\int }_0^4(3t^2+4t+7)dt\\ & d=\left|{[\frac{3t^3}{3}+\frac{4t^2}{2}+7t]}_0^4\right|=\left|{[t^3+2t^2+7t]}_0^4\right|=|\left[\right(4{)}^3+2(4{)}^2+7(4\left)\right]-[0\left]\right|\\ & d=|64+32+28|=124m\\ & a\left(t\right)=V^{\mathrm{\prime }}\left(t\right)=6t+4\Rightarrow a\left(4\right)=6\left(4\right)+4=28\frac{m}{s^2}\end{array}$

(20)- جد المساحة المحددة بالمنحني $f\left(x\right)=(x-1{)}^3$ ومحور السينات في الفترة $[-1,3]$.

$\begin{array}{rl}& (x-1{)}^3=0\Rightarrow x-1=0\Rightarrow x=1\in [-1,3]\\ & \begin{array}{rl}A& =\left|{\int }_{-1}^1\right(x-1{)}^{3}dx|+\left|{\int }_1^3\right(x-1{)}^{3}dx|=\left|{\left[\frac{(x-1{)}^4}{4}\right]}_{-1}^1\right|+\left|{\left[\frac{(x-1{)}^4}{4}\right]}_1^3\right|\\ & =\left|[0-\frac{(-2{)}^4}{4}]\right|+\left|[\frac{(2{)}^4}{4}-0]\right|=-4|+|4\mid =8{\text{unit}}^2\end{array}\end{array}$

(21)- جد الحجم الناتج من دوران المساحة المحصورة بين $y=x^2+1$ والمستقيمين $y=1,y=2$ حول المحور الصادي.

$$\begin{gathered} y=x^2+1\Rightarrow x^2=y-1 \\ v=\pi {\int }_a^b{x}^{2}dy\Rightarrow \pi {\int }_1^2(y-1)dy=\pi {[\frac{y^2}{2}-y]}_1^2 \\ =\pi [\frac{4}{2}-2]-[\frac{1}{2}-1]=\pi (2-2)-(-\frac{1}{2})=\frac{\pi }{2} \left(\mathrm{مكعبة} \mathrm{وحدة}\right) \end{gathered}$$