طباعة الدرس

أتدرب وأحل المسائل

التكامل بالتعويض

أجد كلاً من التكاملات الآتية:

$\int \frac{x}{\sqrt{x^2+4}}dx$ (1)

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

$\int x^2{(2x^3+5)}^{4}dx$ (2)

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

$\int 3x\sqrt{x^2+7}dx$ (3)

$\begin{array}{rl}& \int 3x\sqrt{x^2+7}dx\\ & \begin{array}{rl}u=x^2+7\Rightarrow \frac{du}{dx}& =2x\\ \Rightarrow dx& =\frac{du}{2x}\\ \int 3x\sqrt{x^2+7}dx& =\int 3x\sqrt{u}\times \frac{du}{2x}\\ & =\int \frac{3}{2}{u}^{\frac{1}{2}}du\\ & =u^{\frac{3}{2}}+C\\ & =\sqrt{{(x^2+7)}^3}+C\end{array}\end{array}$

$\int x^6{e}^{1-x^7}dx$ (4)

$\begin{array}{rl}& \int x^6{e}^{1-x^7}dx\\ & u=1-x^7\Rightarrow \frac{du}{dx}=-7x^6\\ & \Rightarrow dx=\frac{du}{-7x^6}\\ & \int x^6{e}^{1-x^7}dx=\int x^6{e}^u\times \frac{du}{-7x^6}\\ & =\int -\frac{1}{7}{e}^{u}du\\ & =-\frac{1}{7}{e}^u+C\\ & =-\frac{1}{7}{e}^{1-x^7}+C\\ & \end{array}$

$\int \frac{x^4}{{(x^5+9)}^3}dx$ (5)

$\begin{array}{rl}& \int \frac{x^4}{{(x^5+9)}^3}dx\\ & u=x^5+9\Rightarrow \frac{du}{dx}=5x^4\\ & \Rightarrow dx=\frac{du}{5x^4}\\ & \int \frac{x^4}{{(x^5+9)}^3}dx=\int \frac{x^4}{u^3}\times \frac{du}{5x^4}\\ & =\int \frac{1}{5}{u}^{-3}du\\ & =-\frac{1}{10}{u}^{-2}+C\\ & =-\frac{1}{10{(x^5+9)}^2}+C\\ & \end{array}$

$\int (3x^2-1)e^{x^3-x}dx$ (6)

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

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

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

$\int \frac{1}{x\ln x}dx$ (8)

$\begin{array}{rl}& \int \frac{1}{x\ln x}dx\\ & u=\ln x\Rightarrow \frac{du}{dx}=\frac{1}{x}\\ & \Rightarrow dx=xdu\\ & \int \frac{1}{x\ln x}dx=\int \frac{1}{xu}\times xdu\\ & =\int \frac{1}{u}du\\ & =\ln \left|u\right|+C\\ & =\ln |\ln x|+C\end{array}$

$\int \sin x(1+\cos x{)}^{4}dx$ (9)

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

$\int {\sin }^{5}2x\cos 2xdx$ (10)

$bbb\begin{array}{rl}& \int {\sin }^{5}2x\cos 2xdx\\ & u=\sin 2x\Rightarrow \frac{du}{dx}=2\cos 2x\\ & \Rightarrow dx=\frac{du}{2\cos 2x}\\ & \int {\sin }^{5}2x\cos 2xdx=\int u^5\cos 2x\times \frac{du}{2\cos 2x}\\ & =\int \frac{1}{2}{u}^{5}du\\ & =\frac{1}{12}{u}^6+C\\ & =\frac{1}{12}(\sin 2x{)}^6+C\\ & \end{array}$

$\int \frac{\sin \left(\frac{1}{x}\right)}{x^2}dx$ (11)

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

$\int \frac{\cos x}{e^{\sin x}}dx$ (12)

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

$\int e^x{(2+e^x)}^{5}dx$ (13)

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

$\int \frac{\cos (\ln x)}{x}dx$ (14)

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

$\int (3x^2-2x-1){(x^3-x^2-x)}^{4}dx$ (15)

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

أجد قيمة كل من التكاملات الآتية:

${\int }_0^2(2x-1)e^{x^2-x}dx$ (16)

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

${\int }_1^2\frac{e^{1/x}}{x^2}dx$ (17)

$\begin{array}{rl}& {\int }_1^2\frac{e^{\frac{1}{x}}}{x^2}dx\\ & u=\frac{1}{x}\Rightarrow \frac{du}{dx}=-\frac{1}{x^2}\\ & \Rightarrow dx=-x^{2}du\\ & x=2\Rightarrow u=\frac{1}{2}\\ & x^2=\frac{1}{x^{\frac{1}{x}}}dx={\int }_1^{\frac{1}{2}}\frac{e^u}{x^2}\times -x^{2}du\\ & ={\int }_1^{\frac{1}{2}}-e^{u}du\\ & =-{e^u|}_1^{\frac{1}{2}}\\ & =-e^{\frac{1}{2}}+e\\ & =-\sqrt{e}+e\end{array}$

${\int }_e^{e^3}\frac{\sqrt{\ln x}}{x}dx$ (18)

$\begin{array}{rl}& {\int }_e^{e^3}\frac{\sqrt{\ln x}}{x}dx\\ & u=\ln x\Rightarrow \frac{du}{dx}=\frac{1}{x}\\ & \Rightarrow dx=xdu\\ & x=e^3\Rightarrow u=\ln e^3=3\\ & x=e\Rightarrow u=\ln e=1\\ & {\int }_e^{e^3}\frac{\sqrt{\ln x}}{x}dx={\int }_1^3\frac{\sqrt{u}}{x}xdu\\ & ={\int }_1^3{u}^{\frac{1}{2}}du\\ & ={\frac{2}{3}{u}^{\frac{3}{2}}|}_1^3\\ & ={\frac{2}{3}\sqrt{u^3}|}_1^3\\ & =\frac{2}{3}\sqrt{3^3}-\frac{2}{3}\sqrt{1^3}\\ & =2\sqrt{3}-\frac{2}{3}\\ & \end{array}$

${\int }_0^1(x^3+x)\sqrt{x^4+2x^2+1}dx$ (19)

$\begin{array}{rl}& {\int }_0^1(x^3+x)\sqrt{x^4+2x^2+1}dx\\ & u=x^4+2x^2+1\Rightarrow \frac{du}{dx}=4x^3+4x\\ & \Rightarrow dx=\frac{du}{4x^3+4x}\\ & \begin{array}{rl}x=1\Rightarrow u=(1{)}^4+2(1{)}^2+1& =4\\ x=0\Rightarrow u=(0{)}^4+2(0{)}^2& +1=1\\ {\int }_0^1(x^3+x)\sqrt{x^4+2x^2+1}dx& ={\int }_1^4(x^3+x)\sqrt{u}\times \frac{du}{4x^3+4x}\\ & ={\int }_1^4(x^3+x)\sqrt{u}\times \frac{du}{4(x^3+x)}\\ & ={\int }_1^4\frac{1}{4}{u}^{\frac{1}{2}}du\\ & ={\frac{1}{6}{u}^{\frac{3}{2}}|}_1^4\\ & ={\frac{1}{6}\sqrt{u^3}|}_1^4\\ & =\frac{1}{6}\sqrt{4^3}-\frac{1}{6}\sqrt{1^3}\\ & =\frac{7}{6}\end{array}\end{array}$

${\int }_0^3\frac{x}{\sqrt{x^2+1}}dx$ (20)

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

${\int }_1^2\frac{2x+1}{{(x^2+x+4)}^3}dx$ (21)

$\begin{array}{rl}& {\int }_1^2\frac{2x+1}{{(x^2+x+4)}^3}dx\\ & u=x^2+x+4\Rightarrow \frac{du}{dx}=2x+1\\ & \Rightarrow dx=\frac{du}{2x+1}\\ & x=2\Rightarrow u=(2{)}^2+2+4=10\\ & x=1\Rightarrow u=(1{)}^2+1+4=6\\ & {\int }_1^2\frac{2x+1}{{(x^2+x+4)}^3}dx={\int }_6^{10}\frac{2x+1}{u^3}\times \frac{du}{2x+1}\\ & ={\int }_6^{10}{u}^{-3}du\\ & =-{\frac{1}{2}{u}^{-2}|}_6^{10}\\ & =-{\frac{1}{2u^2}|}_6^{10}\end{array}$

أجد مساحة المنطقة المظللة في كل من التمثيلين البيانيين الآتيين:

$A=-{\int }_{-1}^{0}6x(x^2+1)dx+{\int }_0^{1}6x(x^2+1)dx$

هناك طريقتان للحل: إما التكامل بالتعويض، أو تكامل كثير حدود بعد توزيع الأقواس.

طريقة التكامل بالتعويض:

$\begin{array}{rl}u& =x^2+1\Rightarrow \frac{du}{dx}=2x\\ \Rightarrow & \Rightarrow dx=\frac{du}{2x}\\ x& =0\Rightarrow u=(0{)}^2+1=1\\ x& =-1\Rightarrow u=(-1{)}^2+1=2\\ x& =1\Rightarrow u=(1{)}^2+1=2\\ A& =-{\int }_{-1}^{0}6x(x^2+1)dx+{\int }_0^{1}6x(x^2+1)dx\\ & =-{\int }_2^{1}6xu\times \frac{du}{2x}+{\int }_1^{2}6xu\times \frac{du}{2x}\\ & =-{\int }_2^{1}3udu+{\int }_1^{2}3udu\\ & =-{\frac{3}{2}{u}^2|}_2^1+{\frac{3}{2}{u}^2|}_1^2\\ & =-\frac{3}{2}(1{)}^2+\frac{3}{2}(2{)}^2+\frac{3}{2}(2{)}^2-\frac{3}{2}(1{)}^2\\ & =9\end{array}$

ومنه مساحة المنطقة المظللة هي 9 وحدات مربعة.

$\begin{array}{rl}& A=-{\int }_{-4}^{0}x\sqrt{16-x^2}dx+{\int }_0^{4}x\sqrt{16-x^2}dx\\ & u=16-x^2\Rightarrow \frac{du}{dx}=-2x\\ & \Rightarrow dx=\frac{du}{-2x}\\ & x=0\Rightarrow u=16-(0{)}^2=16\\ & x=-4\Rightarrow u=16-(-4{)}^2=0\\ & x=4\Rightarrow u=16-(4{)}^2=0\\ & A=-{\int }_{-4}^{0}x\sqrt{16-x^2}dx+{\int }_0^{4}x\sqrt{16-x^2}dx\\ & =-{\int }_0^{16}x\sqrt{u}\times \frac{du}{-2x}+{\int }_{16}^{0}x\sqrt{u}\times \frac{du}{-2x}\\ & ={\int }_0^{16}\frac{1}{2}{u}^{\frac{1}{2}}du+{\int }_{16}^0-\frac{1}{2}{u}^{\frac{1}{2}}du\\ & ={\frac{1}{3}{u}^{\frac{3}{2}}|}_0^{16}+-{\frac{1}{3}{u}^{\frac{3}{2}}|}_{16}^0\\ & ={\frac{1}{3}\sqrt{u^3}|}_0^{16}+-{\frac{1}{3}\sqrt{u^3}|}_{16}^0\\ & =\frac{1}{3}\sqrt{(16{)}^3}-\frac{1}{3}\sqrt{(0{)}^3}-\frac{1}{3}\sqrt{(0{)}^3}+\frac{1}{3}\sqrt{(16{)}^3}\\ & =\frac{128}{3}\end{array}$

ومنه مساحة المنطقة المظللة هي $\frac{128}{3}$ وحدات مربعة.

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

$f^{\mathrm{\prime }}\left(x\right)=x{e}^{4-x^2};(-2,1)$ (24)

$\begin{array}{rl}f\left(x\right)& =\int x{e}^{4-x^2}dx\\ u=& 4-x^2\Rightarrow \frac{du}{dx}=-2x\\ & \Rightarrow dx=\frac{du}{-2x}\\ f\left(x\right)& =\int x{e}^{4-x^2}dx\\ & =\int x{e}^u\frac{du}{-2x}\\ & =\int -\frac{1}{2}{e}^{u}du\\ & =-\frac{1}{2}{e}^u+C\\ & =-\frac{1}{2}{e}^{4-x^2}+C\end{array}$

لإيجاد ثابت التكامل، نعوض النقطة $\left(-2,1\right)$:

$\begin{array}{rl}& f\left(x\right)=-\frac{1}{2}{e}^{4-x^2}+C\Rightarrow f(-2)=-\frac{1}{2}{e}^{4-(-2{)}^2}+C\\ & \Rightarrow 1=-\frac{1}{2}+C\\ & \Rightarrow C=\frac{3}{2}\\ & f\left(x\right)=-\frac{1}{2}{e}^{4-x^2}+\frac{3}{2}\end{array}$

$f^{\mathrm{\prime }}\left(x\right)=\frac{2x}{{(1-x^2)}^2};(0,-1)$ (25)

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

لإيجاد ثابت التكامل، نعوض النقطة $\left(0,-1\right)$:

$\begin{array}{rl}f\left(x\right)=\frac{1}{1-x^2}+C& \Rightarrow f\left(0\right)=\frac{1}{1-0^2}+C\\ & \Rightarrow -1=1+C\\ & \Rightarrow C=-2\\ f\left(x\right)=\frac{1}{1-x^2}-2& \end{array}$

(26) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $\iota \left(t\right)=\frac{-2t}{\sqrt{{(1+t^2)}^3}}$، حيث $t$ الزمن بالثواني، و$v$ سرعته المتجهة بالمتر لكل ثانية. إذا كان الموقع الابتدائي للجسيم $4 m$، فأجد موقع الجسيم بعد $t$ ثانية من بدء الحركة.

$\begin{array}{rl}& s\left(t\right)=\int \frac{-2t}{\sqrt{{(1+t^2)}^3}}dt\\ & u=1+t^2\Rightarrow \frac{du}{dt}=2t\\ & \Rightarrow dt=\frac{du}{2t}\\ & \int \frac{-2t}{\sqrt{{(1+t^2)}^3}}dt=\int \frac{-2t}{\sqrt{u^3}}\times \frac{du}{2t}\\ & =\int -u^{-\frac{3}{2}}du\\ & =2u^{-\frac{1}{2}}+C\\ & =2{(1+t^2)}^{-\frac{1}{2}}+C\\ & =\frac{2}{\sqrt{1+t^2}}+C\end{array}$

بما أن الموقع الابتدائي للجسيم $4 m$، إذن، $s\left(0\right)=4$:

$\begin{array}{rl}& s\left(t\right)=\frac{2}{\sqrt{1+t^2}}+C\Rightarrow f\left(0\right)=\frac{2}{\sqrt{1+0^2}}+C\\ & \Rightarrow 4=2+C\\ & \Rightarrow C=2\\ & s\left(t\right)=\frac{2}{\sqrt{1+t^2}}+2\end{array}$

(27) زراعة: يمثل الاقتران $V\left(t\right)$ سعر دونم أرض زراعية في الأغوار الأردنية (بالدينار) بعد $t$ سنة من الآن. إذا كان: $V^{\mathrm{\prime }}\left(t\right)=\frac{0.4t^3}{\sqrt[3]{0.2t^4+8000}}$ هو معدل التغير في سعر دونم الأرض، فأجد $V\left(t\right)$، علماً بأن سعره الآن $\text{JD }5000$.

$\begin{array}{rl}& V\left(t\right)=\int \frac{0.4t^3}{\sqrt[3]{0.2t^4+8000}}dt\\ & u=0.2t^4+8000\Rightarrow \frac{du}{dt}=0.8t^3\\ & \Rightarrow dt=\frac{du}{0.8t^3}\\ & V\left(t\right)=\int \frac{0.4t^3}{\sqrt[3]{0.2t^4+8000}}dx\\ & =\int \frac{0.4t^3}{\sqrt[3]{u}}\times \frac{du}{0.8t^3}\\ & =\int \frac{1}{2}{u}^{-\frac{1}{3}}du\\ & =\frac{1}{3}{u}^{\frac{2}{3}}+C\\ & =\frac{1}{3}\sqrt[3]{u^2}+C\\ & =\frac{1}{3}\sqrt[3]{{(0.2t^4+8000)}^2}+C\end{array}$

بما أن سعر دونم الأرض الآن هو 5000 دينار، إذن، $V\left(0\right)=5000$ ومنه:

$\begin{array}{rl}& V\left(t\right)=\frac{1}{3}\sqrt[3]{{(0.2t^4+8000)}^2}+C\\ & V\left(0\right)=\frac{1}{3}\sqrt[3]{{\left(0.2\right(0{)}^4+8000)}^2}+C\\ & 5000=\frac{1}{3}\sqrt[3]{(8000{)}^2}+C\\ & 5000=\frac{400}{3}+C\\ & C=\frac{14600}{3}\\ & V\left(t\right)=\frac{1}{3}\sqrt[3]{{(0.2t^4+8000)}^2}+\frac{14600}{3}\end{array}$

(28) سكان: أشارت دراسة إلى أن عدد السكان في إحدى المدن يتغير سنوياً بمعدل يمكن نمذجته بالاقتران: $P^{\mathrm{\prime }}\left(t\right)=\frac{4e^{0.2t}}{\sqrt{4+e^{0.2t}}}$، حيث $t$ عدد السنوات منذ عام 2015 م، و$P\left(t\right)$ عدد السكان بالآلاف. أجد مقدار الزيادة في عدد السكان عام 2015 م إلى عام 2025 م.

$\begin{array}{rl}& \Rightarrow dt=\frac{du}{0.2e^{0.2t}}\\ & t=10\Rightarrow u=4+e^{0.2\left(10\right)}=4+e^2\\ & t=0\Rightarrow u=4+e^{0.2\left(0\right)}=5\\ & {\int }_0^{10}\frac{4e^{0.2t}}{\sqrt{4+e^{0.2t}}}dt={\int }_5^{4+e^2}\frac{4e^{0.2t}}{\sqrt{u}}\times \frac{du}{0.2e^{0.2t}}\\ & ={\int }_5^{4+e^2}20u^{-\frac{1}{2}}du\\ & ={\mathbf{40}{u}^{\frac{1}{2}}|}_5^{4+e^2}\\ & ={40\sqrt{u}|}_5^{4+e^2}\\ & =40\sqrt{4+e^2}-40\sqrt{5}\\ & \approx 46\end{array}$

إذن يزداد عدد سكان هذه المدينة بحوالي 46 ألف شخص من 2015 م إلى 2025 م.