تمارين (5-4)

1- جد $\int \frac{dy}{dx}$ لكل مما يأتي:

- $y=\ln 3x$

$y=\ln 3x\Rightarrow \frac{dy}{dx}=\frac{3}{3x}=\frac{1}{x}$

- $y=\ln \left(\frac{x}{2}\right)$

$y=\ln \left(\frac{x}{2}\right)\Rightarrow \frac{dy}{dx}=\left(\frac{1}{2}\right)\frac{1}{\frac{x}{2}}=\left(\frac{1}{2}\right)\left(\frac{2}{x}\right)=\frac{1}{x}$

- $y=\ln x^2$

$y=\ln x^2\Rightarrow \frac{dy}{dx}=\frac{1}{x^2}\cdot 2x=\frac{2}{x}$

- $y=(\ln x{)}^2$

$y=(\ln x{)}^2\Rightarrow \frac{dy}{dx}=2(\ln x)\cdot \frac{1}{x}=\frac{2}{x}\ln x$

- $y=ln{\left(\frac{1}{x}\right)}^3$

$y=\ln {\left(\frac{1}{x}\right)}^3\Rightarrow y=\ln x^{-3}\Rightarrow \frac{dy}{dx}=\frac{-3x^{-4}}{x^{-3}}=-3x^{-1}=\frac{-3}{x}$

- $y=\ln (2-\cos x)$

$y=\ln (2-\cos x)\Rightarrow \frac{dy}{dx}=\frac{-(-\sin x)}{2-\cos x}=\frac{\sin x}{2-\cos x}$

- $y=e^{-5x^2+3x+5}$

$y=e^{-5x^2+3x+5}\Rightarrow \frac{dy}{dx}=e^{-5x^2+3x+5}\cdot (-10x+3)=(-10x+3)e^{-5x^2+3x+5}$

- $y=9^{\sqrt{x}}$

$y=9^{\sqrt{x}}\Rightarrow \frac{dy}{dx}=9^{\sqrt{x}}\cdot \ln 9\frac{1}{2\sqrt{x}}=\frac{9^{\sqrt{x}}}{2\sqrt{x}}(\ln 9)$

- $y=7^{-\frac{x}{4}}$

$y=7^{-\frac{x}{4}}\Rightarrow \frac{dy}{dx}=7^{-\frac{1}{4}x}\ln 7\cdot (-\frac{1}{4})=7^{-\frac{x}{4}}\left(\frac{-\ln 7}{4}\right)$

- $y=x^2{e}^x$

$y=x^2{e}^x\Rightarrow \frac{dy}{dx}=x^2\cdot e^x+e^x\cdot 2x=x{e}^x(x+2)$

2- جد التكاملات الآتية:

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

مشتقة المقام موجودة فالتكامل $\ln$

$=[\ln |x+1|{]}_0^3=\ln (3+1)-\ln (0+1)=\ln 4-\ln 1=\ln 4-0=\ln 2^2=2\ln 2$

- ${\int }_0^4\frac{2x}{x^2+9}dx$

$\begin{array}{rl}{\int }_0^4\frac{2x}{x^2+9}dx& ={[\ln |x^2+9|]}_0^4=\ln (16+9)-\ln (0+9)=\ln \left(25\right)-\ln \left(9\right)\\ & =\ln 5^2-\ln 3^2=2\ln 5-2\ln 3=2\ln \frac{5}{3}\end{array}$

- ${\int }_{\ln 3}^{\ln 5}{e}^{2x}dx$

$\begin{array}{rl}{\int }_{\ln 3}^{\ln 5}{e}^{2x}dx& =\frac{1}{2}{\int }_{\ln 3}^{\ln 5}{e}^{2x}\left(2\right)dx=\frac{1}{2}{\left[e^{2x}\right]}_{\ln 3}^{\ln 5}=\frac{1}{2}[e^{2\ln 5}-e^{2\ln 3}]\\ & =\frac{1}{2}[e^{\ln 5^2}-e^{\ln 3^2}]=\frac{1}{2}[5^2-3^2]=\frac{1}{2}\left[16\right]=8\end{array}$

- ${\int }_0^{\ln 2}{e}^{-x}dx$

$\begin{array}{rl}{\int }_0^{\ln 2}{e}^{-x}dx& =-{\int }_0^{\ln 2}{e}^{-x}(-1)dx=-{\left[e^{-x}\right]}_0^{\ln 2}=-[e^{-\ln 2}-e^0]\\ & =-[e^{\ln 2^{-1}}-e^0]=-[2^{-1}-1]=-[\frac{1}{2}-1]=\frac{1}{2}\end{array}$

- ${\int }_0^1{(1+e^x)}^2{e}^{x}dx$

${\int }_0^1{(1+e^x)}^2{e}^{x}dx={\left[\frac{{(1+e^x)}^3}{3}\right]}_0^1=[\frac{(1+e{)}^3}{3}-\frac{{(1+e^0)}^3}{3}]=\frac{(1+e{)}^3}{3}-\frac{2^3}{3}=\frac{(1+e{)}^3}{3}-\frac{8}{3}$

ويمكن أن يحل السؤال بطريقة ثانية وهي توزيع $e^x$ على القوس بعد فتح القوس.

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

$\begin{array}{rl}{\int }_0^1\frac{3x^2+4}{(x^3+4x+1)}dx& ={[\ln (x^3+4x+1)]}_0^1=\ln (1^3+4(1)+1-\ln (0^3+4(0)+1)\\ & =\ln 6-\ln 1=\ln 6-0=\ln 6\end{array}$

- ${\int }_1^4\frac{e^{\sqrt{x}}}{2\sqrt{x}}dx$

${\int }_1^4\frac{e^{\sqrt{x}}}{2\sqrt{x}}dx={\int }_1^4{e}^{x^{\frac{1}{2}}\cdot \frac{1}{2}{x}^{-\frac{1}{2}}}dx={\left[e^{x^{\frac{1}{2}}}\right]}_1^4={\left[e^{\sqrt{x}}\right]}_1^4=e^{\sqrt{4}}-e^{\sqrt{1}}=e^2-e^1$

- ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{{\sec }^{2}xdx}{2+\tan x}$

$$\begin{gathered} {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{{\sec }^{2}xdx}{2+\tan x}=[\ln |2+\tan x|{]}_{-\frac{\pi }{4}}^{\frac{\pi }{4}}=\ln (2+\tan \frac{\pi }{4})-\ln (2+\tan (-\frac{\pi }{4})) \\ =\ln (2+1)-\ln (2-1)=\ln 3-\ln 1=\ln 3-0=\ln 3 \end{gathered}$$

- ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{\cos x}{\sqrt{\sin x}}dx$

$\begin{array}{rl}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{\cos x}{\sqrt{\sin x}dx}& ={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{\cos x}{(\sin x{)}^{\frac{1}{2}}}dx={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}(\sin x{)}^{-\frac{1}{2}}\cdot \cos xdx\\ & ={\left[\frac{(\sin x{)}^{\frac{1}{2}}}{\frac{1}{2}}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}=2{\left[\right(\sin x{)}^{\frac{1}{2}}]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}=2[\sqrt{\sin x}{]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}\\ & =2[\sqrt{\sin \frac{\pi }{2}}-\sqrt{\sin \frac{\pi }{6}}]=2[\sqrt{1}-\sqrt{\frac{1}{2}}]=2[1-\frac{1}{\sqrt{2}}]=2-\frac{2}{\sqrt{2}}\end{array}$

- $\int {\cot }^{3}5xdx$

$\begin{array}{rl}& \int {\cot }^{3}5xdx=\int {\cot }^{2}5x\cot 5xdx=\int ({\csc }^{2}5x-1)\cot 5xdx\\ & =\int (\cot 5x{\csc }^{2}5x-\cot 5x)dx=\int (\cot 5x{\csc }^{2}5x)dx-\int (\cot 5x)dx\\ & =\frac{1}{5}\int \cot 5x{\csc }^{2}5x\left(5\right)dx-\frac{1}{5}\int \frac{\cos 5x}{\sin 5x}\left(5\right)dx\\ & =-\frac{1}{5}\frac{{\cot }^{2}5x}{2}-\frac{1}{5}\ln |\sin 5x|+c=-\frac{1}{10}{\cot }^{2}5x-\frac{1}{5}\ln |\sin 5x|+c\end{array}$

- ${\int }_0^{\frac{\pi }{2}}{e}^{\cos x}\sin xdx$

$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}{e}^{\cos x}\sin xdx& =-{\int }_0^{\frac{\pi }{2}}{e}^{\cos x}(-\sin x)dx\\ & =-{\left[e^{\cos x}\right]}_0^{\frac{\pi }{2}}=-[e^{\cos \frac{\pi }{2}}-e^{\cos 0}]=-[e^0-e^1]=-1+e\end{array}$

- ${\int }_1^{2}x{e}^{-\ln x}d$

${\int }_1^{2}x{e}^{-\ln x}dx={\int }_1^{2}x{e}^{\ln x^{-1}}dx={\int }_1^{2}x\cdot x^{-1}dx={\int }_1^{2}dx=[x{]}_1^2=2-1=1$

3- أثبت أن:

- ${\int }_1^8\frac{\sqrt{\sqrt[3]{x}-1}}{\sqrt[3]{x^2}}dx=2$

$\begin{array}{rl}& \mathbf{L}\mathbf{.}\mathbf{H}\mathbf{.}\mathbf{S}={\int }_1^8\frac{{(x^{\frac{1}{3}}-1)}^{\frac{1}{2}}}{{\left(x^2\right)}^{\frac{1}{3}}}dx={\int }_1^8{(x^{\frac{1}{3}}-1)}^{\frac{1}{2}}\cdot x^{-\frac{2}{3}}dx=3{\int }_1^8{(x^{\frac{1}{3}}-1)}^{\frac{1}{2}}\left(\frac{1}{3}{x}^{-\frac{2}{3}}dx\right)\\ & =3{\left[\frac{{(x^{\frac{1}{3}}-1)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_1^8=3\left(\frac{2}{3}\right)[{\left[\right(8{)}^{\frac{1}{3}}-1]}^{\frac{3}{2}}-{\left[\right(1{)}^{\frac{1}{3}}-1]}^{\frac{3}{2}}]\\ & =2\left[\right[2-1{]}^{\frac{3}{2}}-[1-1{]}^{\frac{3}{2}}]=2\left(1\right)=2=\mathbf{R}\mathbf{.}\mathbf{H}\mathbf{.}\mathbf{S}\end{array}$

- ${\int }_{-2}^4|3x-6|dx=30$

$$\begin{gathered} |3x-6|=\{\begin{array}{ll}3x-6& x\ge 2\\ -3x+6& x<2\end{array} \\ \begin{array}{rl}& \mathbf{\text{L.H.S}}={\int }_{-2}^4|3x-6|dx={\int }_{-2}^2(-3x+6)dx+{\int }_2^4(3x-6)dx\\ & ={[\frac{-3}{2}{x}^2+6x]}_{-2}^2+{[\frac{3}{2}{x}^2-6x]}_2^4\\ & =[\frac{-12}{2}+12]-[\frac{-12}{2}-12]+[\frac{48}{2}-24]-[\frac{12}{2}-12]\\ & =[-6+12]-[-6-12]+[24-24]-[6-12]\\ & =6+18+0+6=30=\mathbf{R}\mathbf{.}\mathbf{H}\mathbf{.}\mathbf{S}\end{array} \end{gathered}$$

4- $f\left(x\right)$ دالة مستمرة على الفترة $\left[-2,6\right]$ فإذا كان ${\int }_1^{6}f\left(x\right)dx=6$ وكان ${\int }_{-2}^6\left[f\right(x)+3]dx=32$ فجد ${\int }_{-2}^{1}f\left(x\right)dx$.

$\begin{array}{rl}& \because {\int }_{-2}^6\left[f\right(x)+3]dx=32\\ & {\int }_{-2}^{6}f\left(x\right)dx+{\int }_{-2}^{6}3dx=32\\ & {\int }_{-2}^{6}f\left(x\right)dx+[3x{]}_{-2}^6=32\\ & {\int }_{-2}^{6}f\left(x\right)dx+\left[3\right(6)-3(-2\left)\right]=32\\ & {\int }_{-2}^{6}f\left(x\right)dx+[18+6]=32\\ & {\int }_{-2}^{6}f\left(x\right)dx+\left[24\right]=32\\ & {\int }_{-2}^{6}f\left(x\right)dx=32-24\\ & {\int }_{-2}^{6}f\left(x\right)dx=8\\ & {\int }_{-2}^{6}f\left(x\right)dx={\int }_{-2}^{1}f\left(x\right)dx+{\int }_1^{6}f\left(x\right)dx\\ & {\int }_{-2}^{1}f\left(x\right)dx+6=8\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx=8-6\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx=2\end{array}$

5- إذا علمت أن ${\int }_1^a(x+\frac{1}{2})dx=2{\int }_0^{\frac{\pi }{4}}{\sec }^{2}xdx$ فجد قيمة $a\in R$.

$\begin{array}{rl}& {\int }_1^a(x+\frac{1}{2})dx=2{\int }_0^{\frac{\pi }{4}}{\sec }^{2}xdx\Rightarrow {[\frac{x^2}{2}+\frac{1}{2}x]}_1^a=2[\tan x{]}_0^{\frac{\pi }{4}}\\ & \frac{1}{2}{a}^2+\frac{1}{2}a-[\frac{1}{2}+\frac{1}{2}]=2[\tan \frac{\pi }{4}-\tan 0]\\ & \frac{1}{2}{a}^2+\frac{1}{2}a-1=2[1-0]\Rightarrow \frac{1}{2}{a}^2+\frac{1}{2}a-1=2\\ & \frac{1}{2}{a}^2+\frac{1}{2}a-1-2=0\Rightarrow [\frac{1}{2}{a}^2+\frac{1}{2}a-3=0]\times 2\\ & a^2+a-6=0\Rightarrow (a+3)(a-2)=0\\ & \text{either }a+3=0\Rightarrow a=-3 \left(\mathrm{يهمل}\right)\\ & \text{or}a-2=0\Rightarrow a=2\left(a>1 \left(\mathrm{لأن}\right)\right)\end{array}$

6- لتكن $f\left(x\right)=x^2+2x+k$ حيث $k\in R$ دالة نهايتها الصغرى تساوي $\left(-5\right)$ جد ${\int }_1^{3}f\left(x\right)dx$.

للدالة نهاية صغرى

null

لأنه عندما يأتي في السؤال نهاية صغرى فهي تمثل الإحداثي $y$ في النقطة.

$(-1,-5)$ هي نقطة نهاية صغرى محلية وهي تحقق معادلة الدالة $f\left(x\right)$.

$\begin{array}{rl}& -5=(-1{)}^2+2(-1)+k\Rightarrow -5=1-2+k\Rightarrow k=-4\\ & f\left(x\right)=x^2+2x-4\\ & {\int }_1^{3}f\left(x\right)dx={\int }_1^3(x^2+2x-4)dx\\ & ={[\frac{x^3}{3}+\frac{2x^2}{2}-4x]}_1^3=[\frac{27}{3}+\frac{2\left(9\right)}{2}-12]-[\frac{1}{3}+\frac{2}{2}-4]\\ & =[9+9-12]-[\frac{1}{3}+1-4]=6-\left[\frac{-8}{3}\right]=\frac{18+8}{3}=\frac{26}{3}\end{array}$

7- إذا كان المنحني $f\left(x\right)=(x-3{)}^3+1$ نقطة الانقلاب $(a,b)$ جد القيمة العددية للمقدار null.

نقطة الانقلاب ناتجة من مساواة المشتقة الثانية للصفر.

$$\begin{gathered} f\left(x\right)=(x-3{)}^3+1 \\ \begin{array}{rl}& f\text{'}\left(x\right)=3(x-3{)}^2\\ & f\text{'}\text{'}\left(x\right)=6(x-3)\Rightarrow 6(x-3)=0\stackrel{÷6}{\Rightarrow }x-3=0\Rightarrow x=3\\ & f\left(3\right)=(3-3{)}^3+1=1\left((3,1) \mathrm{هي} \mathrm{الانقلاب} \mathrm{نقطة}\right)\end{array} \\ \begin{array}{rl}& (3,1)\Rightarrow (a,b)\Rightarrow a=3 , b=1\\ & {\int }_0^{b}f\left(x\right)dx-{\int }_0^a\hat{f}\left(x\right)dx\\ & {\int }_0^{1}3(x-3{)}^2-{\int }_0^{3}6(x-3)dx\\ & =3{\left[\frac{(x-3{)}^3}{3}\right]}_0^1-6{\left[\frac{(x-3{)}^2}{2}\right]}_0^3\\ & =\left[\right(1-3{)}^3-(0-3{)}^3]-\left[3\right(3-3{)}^2-3(0-3{)}^2]\\ & =[-8+27]-[0-27]=[19+27]=46\end{array} \end{gathered}$$