إجابات كتاب التمارين

التكامل بالتعويض

أجد كلاً من التكاملات الآتية:

$\int x\sqrt{x^2+3}dx$ (1)

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

$\int x^4{e}^{x^5+2}dx$ (2)

$\begin{array}{rl}& \int x^4{e}^{x^5+2}dx\\ & \begin{array}{rl}& u=x^5+2\Rightarrow \frac{du}{dx}=5x^4\Rightarrow dx=\frac{du}{5x^4}\\ & \int x^4{e}^{x^5+2}dx=\int x^4{e}^u\frac{du}{5x^4}={\int }_0\frac{1}{5}{e}^{u}du\\ & =\frac{1}{5}{e}^u+C=\frac{1}{5}{e}^{x^5+2}+C\end{array}\end{array}$

$\int (x+1){(x^2+2x+5)}^{4}dx$ (3)

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

$\int \frac{(\ln x{)}^3}{x}dx$ (4)

$\begin{array}{rl}& \int \frac{(\ln x{)}^3}{x}dx\\ & u=\ln x\Rightarrow \frac{du}{dx}=\frac{1}{x}\Rightarrow dx=xdu\\ & \begin{array}{rl}\int \frac{(\ln x{)}^3}{x}dx=\int \frac{u^3}{x}xdu& ={\int }_0{u}^{3}du\\ & =\frac{1}{4}{u}^4+C=\frac{1}{4}(\ln x{)}^4+C\end{array}\end{array}$

$\int \frac{\cos x}{{\sin }^{4}x}dx$ (5)

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

$\int \sin x\sqrt{1+3\cos x}dx$ (6)

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

أجد قيمة كل من التكاملات الآتية:

${\int }_1^2\frac{x^2}{{(x^3+1)}^2}dx$ (7)

$\begin{array}{rl}& {\int }_1^2\frac{x^2}{{(x^3+1)}^2}dx\\ & u=x^3+1\Rightarrow \frac{du}{dx}=3x^2\Rightarrow dx=\frac{du}{3x^2}\\ & x=2\Rightarrow u=9\\ & x=1\Rightarrow u=2\\ & {\int }_1^2\frac{x^2}{{(x^3+1)}^2}dx={\int }_2^9\frac{x^2}{u^2}\frac{du}{3x^2}={\int }_2^9\frac{1}{3}{u}^{-2}du\\ & =-{\frac{1}{3u}|}_2^9=-\frac{1}{27}+\frac{1}{6}=\frac{21}{162}\end{array}$

${\int }_0^{1}x\sqrt{3x^2+2}dx$ (8)

$\begin{array}{rl}& {\int }_0^{1}x\sqrt{3x^2+2}dx\\ & u=3x^2+2\Rightarrow \frac{du}{dx}=6x\Rightarrow dx=\frac{du}{6x}\\ & x=1\Rightarrow u=5\\ & x=0\Rightarrow u=2\\ & \begin{array}{rl}{\int }_0^{1}x\sqrt{3x^2+2}dx={\int }_2^{5}x{u}^{\frac{1}{2}}\frac{du}{6x}& ={\int }_2^5\frac{1}{6}{u}^{\frac{1}{2}}du\\ & ={\frac{1}{9}{u}^{\frac{3}{2}}|}_2^5=\frac{1}{9}\sqrt{125}-\frac{1}{9}\sqrt{8}\end{array}\end{array}$

${\int }_e^{e^2}\frac{(\ln x{)}^2}{x}dx$ (9)

$\begin{array}{rl}& {\int }_e^{e^2}\frac{(\ln x{)}^2}{x}dx\\ & u=\ln x\Rightarrow \frac{du}{dx}=\frac{1}{x}\Rightarrow dx=xdu\\ & x=e\Rightarrow u=1\\ & x=e^2\Rightarrow u=2\\ & {\int }_e^{e^2}\frac{(\ln x{)}^2}{x}dx={\int }_1^2\frac{u^2}{x}xdu={\int }_1^2{u}^{2}du\\ & ={\frac{1}{3}{u}^3|}_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}\end{array}$

${\int }_0^1(x+1){(x^2+2x)}^{5}dx$ (10)

$\begin{array}{rl}& {\int }_0^1(x+1){(x^2+2x)}^{5}dx\\ & u=x^2+2x\Rightarrow \frac{du}{dx}=2x+2\Rightarrow dx=\frac{du}{2x+2}\\ & x=1\Rightarrow u=3\\ & x=0\Rightarrow u=0\\ & {\int }_0^1(x+1){(x^2+2x)}^{5}dx={\int }_0^3(x+1)u^5\frac{du}{2x+2}={\int }_0^3\frac{1}{2}{u}^{5}du\\ & ={\frac{1}{12}{u}^6|}_0^3=\frac{729}{12}\end{array}$

(11) أجد مساحة المنطقة المظللة في التمثيل البياني المجاور.

$\begin{array}{rl}& A={\int }_0^{2}x\sqrt{x^2+2}dx\\ & u=x^2+2\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{du}{2x}\\ & x=2\Rightarrow u=6\\ & x=0\Rightarrow u=2\\ & \begin{array}{c}{\int }_0^{2}x\sqrt{x^2+2}dx={\int }_2^{6}x{u}^{\frac{1}{2}}\frac{du}{2x}={\int }_2^6\frac{1}{2}{u}^{\frac{1}{2}}du\\ ={\frac{1}{3}{u}^{\frac{3}{2}}|}_2^6=\frac{1}{3}\sqrt{216}-\frac{1}{3}\sqrt{8}\end{array}\end{array}$

(12) الإيراد الحدي: يمثل الاقتران: $R^{\mathrm{\prime }}\left(x\right)=50+3.5x{e}^{-0.1x^2}$ الإيراد الحدي (بالدينار) لكل قطعة تباع من إنتاج إحدى الشركات، حيث $x$عدد القطع المبيعة، و$R\left(x\right)$ إيراد بيع $x$ قطعة بالدينار. أجد اقتران الإيراد $R\left(x\right)$، علماً بأن $R\left(0\right)=0$. 

$\begin{array}{rl}& R\left(x\right)=\int (50+3.5x{e}^{-0.1x^2})dx=\int 50dx+\int 3.5x{e}^{-0.1x^2}dx\\ & =50x+\int 3.5x{e}^{-0.1x^2}dx\\ & \begin{array}{r}u=-0.1x^2\Rightarrow \frac{du}{dx}=-0.2x\Rightarrow dx=\frac{du}{-0.2x}\end{array}\\ & \begin{array}{rl}& \int (50+3.5x{e}^{-0.1x^2})dx=50x+\int 3.5x{e}^u\frac{du}{-0.2x}\\ & =50x+\int -17.5e^{u}du\end{array}\\ & =50x-17.5e^{-0.1x^2}+C\\ & R\left(0\right)=0\Rightarrow 0-17.5+C=0\Rightarrow C=17.5\end{array}$

يمثل الاقتران ${\mathit{f}}^{\mathbf{\prime }}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ في كل مما يأتي ميل المماس لمنحنى الاقتران $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ المار بالنقطة المعطاة، أستعمل المعلومات المعطاة لإيجاد قاعدة الاقتران $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$:

$f^{\mathrm{\prime }}\left(x\right)=2x{(4x^2-10)}^2;(2,10)$ (13)

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

$f^{\mathrm{\prime }}\left(x\right)=x^2{e}^{-0.2x^3},(0,\frac{3}{2})$ (14)

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

(15) يتحرك جسيم في مسار مستقيم، وتعطى سرعته المتجهة بالاقتران: $v\left(t\right)=\frac{t}{\sqrt{t^2+1}}$، حيث $t$ الزمن بالثواني، و$v$ سرعته المتجهة بالمتر لكل ثانية. إذا بدأ الجسيم حركته من نقطة الأصل، فأجد موقعه بعد $t$ ثانية من بدء الحركة.

$\begin{array}{rl}& s\left(t\right)=\int \frac{t}{\sqrt{t^2+1}}dt\\ & u=t^2+1\Rightarrow \frac{du}{dt}=2t\Rightarrow dt=\frac{du}{2t}\\ & \int \frac{t}{\sqrt{t^2+1}}dt=\int u^{-\frac{1}{2}}\frac{du}{2t}=\int \frac{1}{2}{u}^{-\frac{1}{2}}du=u^{\frac{1}{2}}+C=\sqrt{t^2+1}+C\\ & s\left(t\right)=\sqrt{t^2+1}+C\\ & s\left(0\right)=0\Rightarrow 1+C=0\Rightarrow C=-1\\ & s\left(t\right)=\sqrt{t^2+1}-1\end{array}$