تمارين (1-4)

تمارين (1-4)

(1)- احسب كلاً من التكاملات الآتية:

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

$\begin{array}{rl}{\int }_{-2}^2(3x-2)dx& ={[\frac{3x^2}{2}-2x]}_{-2}^2\\ & =[\frac{3(2{)}^2}{2}-2(2\left)\right]-[\frac{3(-2{)}^2}{2}-2(-2\left)\right]\\ & =[6-4]-[6+4]=2-10=-8\end{array}$

${\int }_1^2(x^{-2}+2x+1)dx$

$\begin{array}{rl}{\int }_1^2(x^{-2}+2x+1)dx& ={[\frac{x^{-1}}{-1}+x^2+x]}_1^2\\ & =[\frac{-1}{2}+(2{)}^2+2]-[\frac{-1}{1}+(1{)}^2+1]\\ & =[\frac{-1}{2}+6]-[-1+1+1]=[\frac{-1}{2}+6]-\left[1\right]=\frac{9}{2}\end{array}$

${\int }_1^3(x^4+4x)dx$

$\begin{array}{rl}{\int }_1^3(x^4+4x)dx& ={[\frac{x^5}{5}+2x^2]}_1^3\\ & =[\frac{(3{)}^5}{5}+2(3{)}^2]-[\frac{1}{5}+2(1{)}^2]\\ & =[\frac{243}{5}+18]-[\frac{1}{5}+2]=\frac{242}{5}+16=\frac{322}{5}\end{array}$

${\int }_0^2|x-1|dx$

$\begin{array}{rl}& {\int }_0^2|x-1|dx\\ & |x-1|=\{\begin{array}{ll}x-1& x\ge 1\\ -x+1& x<1\end{array}\\ & {\int }_0^2|x-1|dx={\int }_0^1(-x+1)dx+{\int }_1^2(x-1)dx\\ & ={[\frac{-x^2}{2}+x]}_0^1+{[\frac{x^2}{2}-x]}_1^2=[\frac{-1}{2}+1]-\left[0\right]+[\frac{4}{2}-2]-[\frac{1}{2}-1]\\ & =\frac{1}{2}+\frac{1}{2}=1\end{array}$

${\int }_{-\frac{\pi }{2}}^0(x+\cos x)dx$

$\begin{array}{rl}{\int }_{-\frac{\pi }{2}}^0(x+\cos x)dx& ={[\frac{x^2}{2}+\sin x]}_{-\frac{\pi }{2}}^0=[\frac{(0{)}^2}{2}+\sin 0]-[\frac{{(-\frac{\pi }{2})}^2}{2}+\sin (-\frac{\pi }{2})]\\ & =\left[0\right]-[\frac{\frac{{\pi }^2}{4}}{2}-\sin \frac{\pi }{2}]=-[\frac{{\pi }^2}{8}-1]=-\frac{{\pi }^2}{8}+1\end{array}$

ملاحظة:

$\sin \frac{\pi }{2}=1 , \sin (-x)=-\sin x$

${\int }_3^2\frac{x^3-1}{x-1}dx$

$\begin{array}{rl}& {\int }_3^2\frac{x^3-1}{x-1}dx\\ & =-{\int }_2^3\frac{x^3-1}{x-1}dx=-{\int }_2^3\frac{(x-1)(x^2+x+1)}{x-1}dx\\ & =-{\int }_2^3(x^2+x+1)dx=-{[\frac{x^3}{3}+\frac{x^2}{2}+x]}_2^3\\ & =-[[\frac{(3{)}^3}{3}+\frac{(3{)}^2}{2}+3]-[\frac{(2{)}^3}{3}+\frac{(2{)}^2}{2}+2]]=-[\frac{27}{3}+\frac{9}{2}+3]-[\frac{8}{3}+\frac{4}{2}+2]\\ & =-\left(\frac{54+27+18-16-12-12}{6}\right)=-\frac{59}{6}\end{array}$

${\int }_1^3\frac{2x^3-4x^2+5}{x^2}dx$

$\begin{array}{rl}{\int }_1^3\frac{2x^3-4x^2+5}{x^2}dx& ={\int }_1^3\frac{2x^3}{x^2}-\frac{4x^2}{x^2}+\frac{5}{x^2}dx={\int }_1^{3}2x-4+5x^{-2}dx\\ & ={[x^2-4x+\frac{5x^{-1}}{-1}]}_1^3={[x^2-4x-\frac{5}{x}]}_1^3\\ & =[9-12-\frac{5}{3}]-[1-4-5]=[-3-\frac{5}{3}]-8\\ & =-\frac{14}{3}+8=\frac{-14+24}{3}=\frac{10}{3}\end{array}$

(2)- أثبت أن $F\left(x\right)$ هي دالة مقابلة للدالة $f\left(x\right)$ حيث $F:[0,\frac{\pi }{6}]\to R$ حيث $F\left(x\right)=\sin x+x$

$f\left(x\right)=1+\cos x$ حيث $f:[0,\frac{\pi }{6}]\to R$ ثم أحسب ${\int }_0^{\frac{\pi }{6}}f\left(x\right)dx$.

لكي نثبت أن $F\left(x\right)$ دالة مقابلة للدالة $f\left(x\right)$

$\begin{array}{rl}& F\left(x\right)=\sin x+x\\ & F\left(x\right)=\cos x+1\\ & F\left(x\right)=\cos x+1=f\left(x\right)\end{array}$

الدالة $F\left(x\right)$ هي دالة مقابلة للدالة $f\left(x\right)$

${\int }_0^{\frac{\pi }{6}}f\left(x\right)dx=F\left(\frac{\pi }{6}\right)-F\left(0\right)=\sin \frac{\pi }{6}+\frac{\pi }{6}-[\sin 0-0]=\frac{1}{2}+\frac{\pi }{6}=\frac{3+\pi }{6}$

(3)- أوجد كلاً من التكاملات الآتية:

${\int }_1^4(x-2)(x+1{)}^{2}dx$

$\begin{array}{rl}& {\int }_1^4(x-2)(x+1{)}^{2}dx={\int }_1^4(x-2)(x^2+2x+1)dx\\ & ={\int }_1^4(x^3+2x^2+x-2x^2-4x-2)dx={\int }_1^4(x^3-3x-2)dx\\ & {=[\frac{x^4}{4}-\frac{3}{2}{x}^2-2x]}_1^4=[\frac{4^4}{4}-\frac{3}{2}(16)-8]-[\frac{1}{4}-\frac{3}{2}-2]\end{array}$

${\int }_{-1}^1|x+1|dx$

$$\begin{gathered} {\int }_{-1}^1|x+1|dx \\ |x+1|=\{\begin{array}{cc}x+1& x\ge -1\\ -x-1& x<-1 \left(\mathrm{الفترة} \mathrm{خارج}\right)\end{array} \end{gathered}$$

لذا في هذه الحالة نأخذ الجزء الموجب فقط.

${\int }_{-1}^1|x+1|dx={\int }_{-1}^1(x+1)dx={[\frac{x^2}{2}+x]}_{-1}^1=[\frac{1}{2}+1]-[\frac{1}{2}-1]=\frac{3}{2}+\frac{1}{2}=2$

${\int }_2^3\frac{x^4-1}{x-1}dx$

$\begin{array}{rl}& {\int }_2^3\frac{x^4-1}{x-1}dx\\ & ={\int }_2^3\frac{(x^2-1)(x^2+1)}{x-1}dx={\int }_2^3\frac{(x-1)(x+1)(x^2+1)}{x-1}dx\\ & ={\int }_2^3(x+1)(x^2+1)dx={\int }_2^3(x^3+x+x^2+1)dx\\ & ={[\frac{x^4}{4}+\frac{x^2}{2}+\frac{x^3}{3}+x]}_2^3=[\frac{3^4}{4}+\frac{3^2}{2}+\frac{3^3}{3}+3]-[\frac{16}{4}+\frac{4}{2}+\frac{8}{3}+2]\\ & =[\frac{81}{4}+\frac{9}{2}+12]-[8+\frac{8}{3}]=\frac{81}{4}+\frac{9}{2}+12-8-\frac{8}{3}=\frac{81}{4}+\frac{9}{2}+4-\frac{8}{3}\\ & =\frac{243+54+48-32}{12}=\frac{313}{12}\end{array}$

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

$\begin{array}{rl}& {\int }_0^1\sqrt{x}(\sqrt{x}+2{)}^{2}dx={\int }_0^1\sqrt{x}(x+4\sqrt{x}+4)dx\\ & ={\int }_0^1(x\sqrt{x}+4x+4\sqrt{x})dx={\int }_0^1\left(x\right(x{)}^{\frac{1}{2}}+4x+4x^{\frac{1}{2}})dx\\ & ={\int }_0^1\left(\right(x{)}^{\frac{3}{2}}+4x+4x^{\frac{1}{2}})dx={[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+\frac{4x^2}{2}+\frac{4x^{\frac{3}{2}}}{\frac{3}{2}}]}_0^1\\ & =[\frac{2}{5}+2+\frac{8}{3}]-\left[0\right]=\frac{6+30+40}{15}=\frac{76}{15}\end{array}$

(4)- إذا كانت $f\left(x\right)=\{\begin{array}{ll}2x& \mathrm{\forall }x\ge 3\\ 6& \mathrm{\forall }x<3\end{array}$ جد ${\int }_1^{4}f\left(x\right)dx$

لنبرهن أن الدالة $f\left(x\right)$ مستمرة على الفترة $\left[1,4\right]$

$$\begin{gathered} \text{1) }f\left(3\right)=2\left(3\right)=6 \left(\mathrm{معرفة}\right) \\ 2) \underset{x\to 3}{lim}f(x)=\{\begin{array}{l}\underset{x\to 3^{+}}{lim}2x=6=L_1\\ \underset{x\to 3^{-}}{lim}6=6=L_2\end{array} \\ {L}_1=L_2 \left(\mathrm{موجودة} \mathrm{الغاية}\right) \\ \text{3) }\underset{x\to 3}{lim}f(x)=f(x)=6 \left(\mathrm{مستمرة} \mathrm{الدالة}\right) \\ \begin{array}{rl}{\int }_1^{4}f\left(x\right)dx& ={\int }_1^{3}f\left(x\right)dx+{\int }_3^{4}f\left(x\right)dx={\int }_1^{3}6dx+{\int }_3^{4}2xdx\\ & =[6x{]}_1^3+{\left[x^2\right]}_3^4=[18-6]+[16-9]=12+7=19\end{array} \end{gathered}$$

(5)- إذا كانت $f\left(x\right)=\{\begin{array}{cc}3x^2& \mathrm{\forall }x\ge 0\\ 2x& \mathrm{\forall }x<0\end{array}$ فأوجد ${\int }_{-1}^{3}f\left(x\right)$

نبرهن أن الدالة مستمرة على الفترة $\left[-1,3\right]$

$$\begin{gathered} \text{1) }f\left(0\right)=3(0{)}^2=0 \\ 2) \underset{x\to 0}{lim}f\left(x\right)=\{\begin{array}{l}\underset{x\to 0^{+}}{lim}3x^2=3(0{)}^2=0=L_1\\ \underset{x\to 3^{-}}{lim}2x=2\left(0\right)=0=L_2\end{array} \\ {L}_1=L_2 \left(\mathrm{موجودة} \mathrm{الغاية}\right) \\ \text{3) }\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)=0\begin{array}{}\\ \end{array} \\ {\int }_{-1}^{3}f\left(x\right)={\int }_{-1}^{0}2x+{\int }_0^{3}3x^2={\left[x^2\right]}_{-1}^0+{\left[x^3\right]}_0^3=[0-1]+[27-0]=-1+27=26 \end{gathered}$$