NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.1

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.1 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

NCERT Question-Answer

Class 12 Mathematics

Chapter-7 (Integrals)

Questions and answers given in practice

Chapter-7 (Integrals)

Exercise 7.1

Q1. Find an anti-derivative (or integral) of the following functions by the method of inspection. sin 2x

Answer. The anti derivative of sin 2x is a function of x whose derivative is sin 2x. It is known that, $\begin{array}{c}\frac{d}{dx}\left(\cos 2x\right)=-2\sin 2x\\ \Rightarrow \sin 2x=-\frac{1}{2}\frac{d}{dx}\left(\cos 2x\right)\\ \therefore \sin 2x=\frac{d}{dx}(-\frac{1}{2}\cos 2x)\end{array}$ Therefore, the anti derivative of $\sin 2x\text{is}-\frac{1}{2}\cos 2x$

Q2. Find an anti derivative (or integral) of the following functions by the method of inspection. cos 3x

Answer.  The anti derivative of cos 3x is a function of x whose derivative is cos 3x. It is known that, $\begin{array}{c}\frac{d}{dx}\left(\sin 3x\right)=3\cos 3x\\ \Rightarrow \cos 3x=\frac{1}{3}\frac{d}{dx}\left(\sin 3x\right)\\ \therefore \cos 3x=\frac{d}{dx}\left(\frac{1}{3}\sin 3x\right)\end{array}$ Therefore, the anti derivative of $\cos 3x\text{is}\frac{1}{3}\sin 3x$.

Q3. Find an anti derivative (or integral) of the following functions by the method of inspection. $e^{2x}$

Answer. The anti derivative of $e^{2x}$ is the function of x whose derivative is $e^{2x}$. It is known that, $\begin{array}{c}\frac{d}{dx}\left(e^{2x}\right)=2e^{2x}\\ \Rightarrow e^{2x}=\frac{1}{2}\frac{d}{dx}\left(e^{2x}\right)\\ \therefore e^{2x}=\frac{d}{dx}\left(\frac{1}{2}{e}^{2x}\right)\end{array}$ Therefore, the anti derivative of $e^{2x}\text{is}\frac{1}{2}{e}^{2x}$

Q4. Find an anti derivative (or integral) of the following functions by the method of inspection. $(ax+b{)}^2$

Answer. The anti derivative of $(ax+b{)}^2$ is the function of x whose derivative is $(ax+b{)}^2$. It is known that $\begin{array}{c}\frac{d}{dx}(ax+b{)}^3=3a(ax+b{)}^2\\ \Rightarrow (ax+b{)}^2=\frac{1}{3a}\frac{d}{dx}(ax+b{)}^3\\ \therefore (ax+b{)}^2=\frac{d}{dx}\left(\frac{1}{3a}\right(ax+b{)}^3)\end{array}$ Therefore, the anti derivative of $(ax+b{)}^2\text{is}\frac{1}{3a}(ax+b{)}^3$

Q5. Find an anti derivative (or integral) of the following functions by the method of inspection: 

Answer. The anti derivative of $\sin 2x-4e^{3x}$ is the function of x whose derivative is $\sin 2x-4e^{3x}$. It is known that, $\frac{d}{dx}(-\frac{1}{2}\cos 2x-\frac{4}{3}{e}^{3x})=\sin 2x-4e^{3x}$ Therefore, the anti derivative of $(\sin 2x-4e^{3x})\text{is}(-\frac{1}{2}\cos 2x-\frac{4}{3}{e}^{3x})$.

Q6. Find the following integral $\int (4e^{3x}+1)dx$

Answer. $\begin{array}{c}\int (4e^{3x}+1)dx\\ =4\int e^{3x}dx+\int 1dx\\ =4\left(\frac{e^{3x}}{3}\right)+x+C\\ =\frac{4}{3}{e}^{3x}+x+C\end{array}$

Q7. Find the following integral $\int x^2(1-\frac{1}{x^2})dx$

Answer. $\begin{array}{c}\int x^2(1-\frac{1}{x^2})dx\\ =\int (x^2-1)dx\\ =\int x^{2}dx-\int 1dx\\ =\frac{x^3}{3}-x+C\end{array}$

Q8. Find the following integral $\int (a{x}^2+bx+c)dx$

Answer. $\begin{array}{cc}& \int (a{x}^2+bx+c)dx\\ & =a\int x^{2}dx+b\int xdx+c\int 1.dx\\ & =a\left(\frac{x^3}{3}\right)+b\left(\frac{x^2}{2}\right)+cx+C\\ & =\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+C\end{array}$

Q9. Find the following integral $\int (2x^2+e^x)dx$

Answer. $\begin{array}{c}\int (2x^2+e^x)dx\\ =2\int x^{2}dx+\int e^{x}dx\\ =2\left(\frac{x^3}{3}\right)+e^x+C\\ =\frac{2}{3}{x}^3+e^x+C\end{array}$

Q10. Find the following integral $\int {(\sqrt{x}-\frac{1}{\sqrt{x}})}^{2}dx$

Answer. $\begin{array}{cc}& \int {(\sqrt{x}-\frac{1}{\sqrt{x}})}^{2}dx\\ & =\int (x+\frac{1}{x}-2)dx\\ & =\int xdx+\int \frac{1}{x}dx-2\int 1dx\\ & =\frac{x^2}{2}+\log |x|-2x+C\end{array}$

Q11. Find the following integral $\int \frac{x^3+5x^2-4}{x^2}dx$

Answer. $\begin{array}{c}\int \frac{x^3+5x^2-4}{x^2}dx\\ =\int (x+5-4x^{-2})dx\\ =\int xdx+5\int 1\cdot dx-4\int x^{-2}dx\\ =\frac{x^2}{2}+5x-4\left(\frac{x^{-1}}{-1}\right)+C\\ =\frac{x^2}{2}+5x+\frac{4}{x}+C\end{array}$

Q12. Find the following integral $\int \frac{x^3+3x+4}{\sqrt{x}}dx$

Answer. $\begin{array}{c}\int \frac{x^3+3x+4}{\sqrt{x}}dx\\ =\int (x^{\frac{5}{2}}+3x^{\frac{1}{2}}+4x^{\frac{1}{2}})dx\\ =\frac{x^7}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\frac{4\left(\frac{1}{x^2}\right)}{\frac{1}{2}}+C\\ =\frac{2}{7}{x}^{\frac{7}{2}}+2x^{\frac{3}{2}}+8x^{\frac{1}{2}}+C\\ =\frac{2}{7}{x}^2+2x^{\frac{3}{2}}+8\sqrt{x}+C\end{array}$

Q13. Find the following integral $\int \frac{x^3-x^2+x-1}{x-1}dx$

Answer. $\begin{array}{c}\int \frac{x^3-x^2+x-1}{x-1}dx\\ \text{On dividing, we obtain}\\ =\int (x^2+1)dx\\ =\int x^{2}dx+\int 1dx\\ =\frac{x^3}{3}+x+C\end{array}$

Q14. Find the following integral $\int (1-x)\sqrt{x}dx$

Answer. $\begin{array}{c}\int (1-x)\sqrt{x}dx\\ =\int (\sqrt{x}-x^{\frac{3}{2}})dx\\ =\int x^{\frac{1}{2}}dx-\int x^{\frac{3}{2}}dx\\ =\frac{x^{\frac{3}{2}}}{2}-\frac{x^2}{2}+C\\ =\frac{2}{3}{x}^2-\frac{2}{5}{x}^2+C\end{array}$

Q15. Find the following integral $\int \sqrt{x}(3x^2+2x+3)dx$

Answer. $\begin{array}{c}\int \sqrt{x}(3x^2+2x+3)dx\\ =\int (3x^{\frac{5}{2}}+2x^{\frac{3}{2}}+3x^{\frac{1}{2}})dx\\ =3\int x^{\frac{5}{2}}dx+2\int x^{\frac{3}{2}}dx+3\int x^{\frac{1}{2}}dx\end{array}$ $\begin{array}{c}=3\left(\frac{x^{\frac{7}{2}}}{\frac{7}{2}}\right)+2\left(\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right)+3\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\\ =\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C\end{array}$

Q16. Find the following integral $\int (2x-3\cos x+e^x)dx$

Answer. $\begin{array}{c}\int (2x-3\cos x+e^x)dx\\ =2\int xdx-3\int \cos xdx+\int e^{x}dx\\ =\frac{2x^2}{2}-3\left(\sin x\right)+e^x+C\\ =x^2-3\sin x+e^x+C\end{array}$

Q17. Find the following integral $\int (2x^2-3\sin x+5\sqrt{x})dx$

Answer. $\int (2x^2-3\sin x+5\sqrt{x})dx$ $\begin{array}{c}=2\int x^{2}dx-3\int \sin xdx+5\int x^{\frac{1}{2}}dx\\ =\frac{2x^3}{3}-3(-\cos x)+5\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\\ =\frac{2}{3}{x}^3+3\cos x+\frac{10}{3}{x}^2+C\end{array}$

Q18. Find the following integral $\int \sec x(\sec x+\tan x)dx$

Answer. $\begin{array}{c}\int \sec x(\sec x+\tan x)dx\\ =\int ({\sec }^{2}x+\sec x\tan x)dx\\ =\int {\sec }^{2}xdx+\int \sec x\tan xdx\\ =\tan x+\sec x+C\end{array}$

Q19. Find the following integral $\int \frac{{\sec }^{2}x}{{\csc }^{2}x}dx$

Answer. $\int \frac{{\sec }^{2}x}{{\csc }^{2}x}dx$ $\begin{array}{cc}& =\int \frac{{\cos }^{2}x}{\int \frac{1}{{\sin }^{2}x}}dx\\ & =\int \frac{{\sin }^{2}x}{{\cos }^{2}x}dx\\ & =\int {\tan }^{2}x\\ & =\int ({\sec }^{2}x-1)dx\\ & =\int {\sec }^{2}xdx\\ & =\tan x-x+C\end{array}$

Q20. Find the following integral $\int \frac{2-3\sin x}{{\cos }^{2}x}dx$

Answer. $\begin{array}{cc}& \int \frac{2-3\sin x}{{\cos }^{2}x}dx\\ & =\int (\frac{2}{{\cos }^{2}x}-\frac{3\sin x}{{\cos }^{2}x})dx\\ & =\int 2{\sec }^{2}xdx-3\int \tan x\sec xdx\\ & =2\tan x-3\sec x+C\end{array}$

Q21. The anti derivative of $(\sqrt{x}+\frac{1}{\sqrt{x}})$ equals. (A) $\frac{1}{3}{x}^{\frac{1}{3}}+2x^{\frac{1}{2}}+C\text{(B)}\frac{2}{3}{x}^{\frac{2}{3}}+\frac{1}{2}{x}^2+C$ (C) $\frac{2}{3}{x}^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\text{(D)}\frac{3}{2}{x}^{\frac{3}{2}}+\frac{1}{2}{x}^{\frac{1}{2}}+C$

Answer. $\begin{array}{c}(\sqrt{x}+\frac{1}{\sqrt{x}})dx\\ =\int x^{\frac{1}{2}}dx+\int x^{\frac{1}{2}}dx\\ =\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C\\ =\frac{2}{3}{x}^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\end{array}$ Hence, the correct Answer is C.

Q22. If $\frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}\text{such that}f\left(2\right)=0.\text{Then}f\left(x\right)$ (A) $x^4+\frac{1}{x^3}-\frac{129}{8}\text{(B)}{x}^3+\frac{1}{x^4}+\frac{129}{8}$ (C) $x^4+\frac{1}{x^3}+\frac{129}{8}\text{(D)}{x}^3+\frac{1}{x^4}-\frac{129}{8}$

Answer. $\begin{array}{c}\text{It is given that,}\\ \frac{d}{dx}f\left(x\right)=4x^3-\frac{3}{x^4}\\ \therefore {\text{Anti derivative of}}^{4x^3-\frac{3}{x^4}}=f\left(x\right)\end{array}$ $\begin{array}{c}\therefore f\left(x\right)=\int 4x^3-\frac{3}{x^4}dx\\ f\left(x\right)=4\int x^{3}dx-3\int \left(x^{-4}\right)dx\\ f\left(x\right)=4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C\\ f\left(x\right)=x^4+\frac{1}{x^3}+C\end{array}$ $\begin{array}{c}\text{Also,}\\ f\left(2\right)=0\\ \therefore f\left(2\right)=(2{)}^4+\frac{1}{(2{)}^3}+C=0\\ \Rightarrow 16+\frac{1}{8}+C=0\\ \Rightarrow C=-(16+\frac{1}{8})\\ \Rightarrow C=\frac{-129}{8}\end{array}$ $\begin{array}{c}\therefore f\left(x\right)=x^4+\frac{1}{x^3}-\frac{129}{8}\\ \text{Hence, the correct Answer is A.}\end{array}$

Chapter-7 (integrals)

• Exercise 7.2
• Exercise 7.3
• Exercise 7.4
• Exercise 7.5
• Exercise 7.6
• Exercise 7.7
• Exercise 7.8
• Exercise 7.9
• Exercise 7.10
• Exercise 7.11