NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.9

Exercise 7.9

Q1. Evaluate the definite integral : ${\int }_{-1}^1(x+1)dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_{-1}^4(x+1)dx\\ \int (x+1)dx=\frac{x^2}{2}+x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $I=F\left(1\right)-F(-1)$ $\begin{array}{c}=(\frac{1}{2}+1)-(\frac{1}{2}-1)\\ =\frac{1}{2}+1-\frac{1}{2}+1\\ =2\end{array}$

Q2. Evaluate the definite integral : ${\int }_x^1\frac{1}{x}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^3\frac{1}{x}dx\\ \int \frac{1}{x}dx=\log |x|=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(3\right)-F\left(2\right)\\ & =\log |3|-\log |2|=\log \frac{3}{2}\end{array}$

Q3. Evaluate the definite integral : ${\int }^2(4x^3-5x^2+6x+9)dx$

Answer. $\begin{array}{cc}\text{Let}I=\int (4x^3-5x^2+6x+9)dx& \\ \int (4x^3-5x^2+6x+9)dx& =4\left(\frac{x^4}{4}\right)-5\left(\frac{x^3}{3}\right)+6\left(\frac{x^2}{2}\right)+9\left(x\right)\\ & =x^4-\frac{5x^3}{3}+3x^2+9x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(2\right)-F\left(1\right)\\ I& =\{2^4-\frac{5\cdot (2{)}^3}{3}+3(2{)}^2+9\left(2\right)\}-\left\{\right(1{)}^4-\frac{5(1{)}^3}{3}+3(1{)}^2+9(1\left)\right\}\\ & =(16-\frac{40}{3}+12+18)-(1-\frac{5}{3}+3+9)\\ & =16-\frac{40}{3}+12+18-1+\frac{5}{3}-3-9\\ & =33-\frac{35}{3}\\ & =\frac{99-35}{3}\\ & =\frac{64}{3}\end{array}$

Q4. Evaluate the definite integral : ${\int }_0^x\sin 2xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{\pi }\sin 2xdx\\ \int \sin 2xdx=\left(\frac{-\cos 2x}{2}\right)=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(0\right)\\ & =-\frac{1}{2}[\cos 2\left(\frac{1}{4}\right)-\cos 0]\\ & =-\frac{1}{2}[\cos \left(\frac{-}{2}\right)-\cos 0]\\ & =-\frac{1}{2}[0-1]\\ & =\frac{1}{2}\end{array}$

Q5. Evaluate the definite integral : ${\int }^{\frac{x}{2}}\cos 2xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^{\frac{\pi }{2}}\cos 2xdx\\ \int \cos 2xdx=\left(\frac{\sin 2x}{2}\right)=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{2}\right)-F\left(0\right)\\ & =\frac{1}{2}[\sin 2\left(\frac{\pi }{2}\right)-\sin 0]\\ & =\frac{1}{2}[\sin \pi -\sin 0]\\ & =\frac{1}{2}[0-0]=0\end{array}$

Q6. Evaluate the definite integral : ${\int }_1^6{e}^{x}dx$

Answer. $\begin{array}{c}\text{Let}I={\int }_4^6{e}^{x}dx\\ \int e^{x}dx=e^x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(5\right)-F\left(4\right)\\ & =e^5-e^4\\ & =e^4(e-1)\end{array}$

Q7. Evaluate the definite integral : ${\int }_0^x\tan xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_5^{\pi }\tan xdx\\ \int \tan xdx=-\log |\cos x|=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(0\right)\\ & =-\log \left|\cos \frac{\pi }{4}\right|+\log |\cos 0|\\ & =-\log |\frac{1}{\sqrt{2}|}+\log |l|\\ & =-\log (2{)}^{-\frac{1}{2}}\\ & =\frac{1}{2}\log 2\end{array}$

Q8. Evaluate the definite integral : ${\int }_6^{\frac{\pi }{4}}\cos \mathrm{ec}xdx$

Answer. $I={\int }_{\pi }^{\frac{\pi }{4}}\cos \mathrm{ec}xdx$ $\int \csc xdx=\log |\csc x-\cot x|=F\left(x\right)$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(\frac{\pi }{4}\right)-F\left(\frac{\pi }{6}\right)\\ & =\log |\csc \frac{\pi }{4}-\cot \frac{\pi }{4}|-\log |\csc \frac{\pi }{6}-\cot \frac{\pi }{6}|\\ & =\log |\sqrt{2}-1|-\log |2-\sqrt{3}|\\ & =\log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\end{array}$

Q9. Evaluate the definite integral : ${\int }_0^4\frac{dx}{\sqrt{1-x^2}}$

Answer. $\begin{array}{c}\text{Let}I={\int }_0^4\frac{dx}{\sqrt{1-x^2}}\\ \int \frac{dx}{\sqrt{1-x^2}}={\sin }^{-1}x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & ={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)\\ & =\frac{\pi }{2}-0\\ & =\frac{\pi }{2}\end{array}$

Q10. Evaluate the definite integral : $\int \frac{dx}{1+x^2}$

Answer. $\begin{array}{c}\text{Let}I=\int \frac{dx}{1+x^2}\\ \int \frac{dx}{1+x^2}={\tan }^{-1}x=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(1\right)-F\left(0\right)\\ & ={\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)\\ & =\frac{\pi }{4}\end{array}$

Q11. Evaluate the definite integral : ${\int }_2^3\frac{dx}{x^2-1}$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^3\frac{dx}{x^2-1}\\ \int \frac{dx}{x^2-1}=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(3\right)-F\left(2\right)\\ & =\frac{1}{2}[\log \left|\frac{3-1}{3+1}\right|-\log |\frac{2-1}{2+1}]\\ & =\frac{1}{2}[\log \left|\frac{2}{4}\right|-\log \frac{1}{3}]\\ & =\frac{1}{2}[\log \frac{1}{2}-\log \frac{1}{3}]\\ & =\frac{1}{2}\left[\log \frac{3}{2}\right]\end{array}$

Q12. Evaluate the definite integral : ${\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx$

Answer. $\begin{array}{c}\text{Let}I={\int }_1^{\frac{\pi }{2}}{\cos }^{2}xdx\\ \int {\cos }^{2}xdx=\int \left(\frac{1+\cos 2x}{2}\right)dx=\frac{x}{2}+\frac{\sin 2x}{4}=\frac{1}{2}(x+\frac{\sin 2x}{2})=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =[F\left(\frac{\pi }{2}\right)-F(0\left)\right]\\ & =\frac{1}{2}[(\frac{\pi }{2}-\frac{\sin \pi }{2})-(0+\frac{\sin 0}{2})]\\ & =\frac{1}{2}[\frac{\pi }{2}+0-0-0]\\ & =\frac{\pi }{4}\end{array}$

Q13. Evaluate the definite integral : ${\int }_2^3\frac{xdx}{x^2+1}$

Answer. $\begin{array}{c}\text{Let}I={\int }_2^3\frac{x}{x^2+1}dx\\ \int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{2x}{x^2+1}dx=\frac{1}{2}\log (1+x^2)=F\left(x\right)\end{array}$ By second fundamental theorem of calculus , we obtain $\begin{array}{cc}I& =F\left(3\right)-F\left(2\right)\\ & =\frac{1}{2}[\log (1+(3{)}^2)-\log (1+(2{)}^2)]\\ & =\frac{1}{2}\left[\log \right(10)-\log (5\left)\right]\\ & =\frac{1}{2}\log \left(\frac{10}{5}\right)=\frac{1}{2}\log 2\end{array}$

Q14. Evaluate the definite integral : $\int \frac{2x+3}{5x^2+1}dx$

Answer. $I={\int }_0^4\frac{2x+3}{5x^2+1}dx$ $\begin{array}{cc}\int \frac{2x+3}{5x^2+1}dx& =\frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx\\ & =\frac{1}{5}\int \frac{10x+15}{5x^2+1}\end{array}$