NCERT Solutions for Exercise 7.1 Class 12 Maths Chapter 7 - Integrals

Question 2: Find an anti derivative (or integral) of the function by the method of inspection. $\cos 3x$

Answer:

Given $\cos 3x$ ;

So, the antiderivative of $\cos 3x$ is a function of x whose derivative is $\cos 3x$ .

$\frac{d}{dx}(\sin 3x)=3\cos 3x$

$⟹\cos 3x=\frac{1}{3}\frac{d}{dx}(\sin 3x)$

Therefore, we have the anti derivative of $\cos 3x$ is $\frac{1}{3}\sin 3x$ .

Question 3: Find an anti derivative (or integral) of the function by the method of inspection. $e^{2x}$

Answer:

Given $e^{2x}$ ;

So, the anti derivative of $e^{2x}$ is a function of x whose derivative is $e^{2x}$ .

$\frac{d}{dx}\left(e^{2x}\right)=2e^{2x}$

$⟹e^{2x}=\frac{1}{2}\frac{d}{dx}(e^{2x})$

$\therefore e^{2x}=\frac{d}{dx}(\frac{1}{2}{e}^{2x})$

Therefore, we have the anti derivative of $e^{2x}$ is $\frac{1}{2}{e}^{2x}$ .

Question 4: Find an anti derivative (or integral) of the function by the method of inspection. $(ax+b{)}^2$

Answer:

Given $(ax+b{)}^2$ ;

So, the anti derivative of $(ax+b{)}^2$ is a function of x whose derivative is $(ax+b{)}^2$ .

$\frac{d}{dx}(ax+b{)}^3=3a(ax+b{)}^2$

$\Rightarrow (ax+b{)}^2=\frac{1}{3a}\frac{d}{dx}(ax+b{)}^3$

$\therefore (ax+b{)}^2=\frac{d}{dx}[\frac{1}{3a}(ax+b{)}^3]$

Therefore, we have the anti derivative of $(ax+b{)}^2$ is $[\frac{1}{3a}(ax+b{)}^3]$ .

Question 5: Find an anti derivative (or integral) of the function by the method of inspection. $\sin 2x-4e^{3x}$

Answer:

Given $\sin 2x-4e^{3x}$ ;

So, the anti derivative of

$\frac{d}{dx}(-\frac{1}{2}\cos 2x-\frac{4}{3}{e}^{3x})=\sin 2x-4e^{3x}$

Therefore, we have the anti derivative of $\sin 2x-4e^{3x}$ is $(-\frac{1}{2}\cos 2x-\frac{4}{3}{e}^{3x})$ .

Question 6: Find the integral$\int (4e^{3x}+1)dx$

Answer:

Given intergral $\int (4e^{3x}+1)dx$ ;

$4\int e^{3x}dx+\int 1dx=4\left(\frac{e^{3x}}{3}\right)+x+C$

or $\frac{4}{3}{e}^{3x}+x+C$ , where C is any constant value.

Question 7: Find the integral$\int x^2(1-\frac{1}{x^2})dx$

Answer:

Given intergral $\int x^2(1-\frac{1}{x^2})dx$ ;

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

or $\frac{x^3}{3}-x+C$ , where C is any constant value.

Question 8: Find the integral $\int (a{x}^2+bx+c)dx$

Answer:

Given intergral $\int (a{x}^2+bx+c)dx$ ;

$\int a{x}^{2}dx+\int bxdx+\int cdx$

$=a\int x^{2}dx+b\int xdx+c\int dx$

$=a\frac{x^3}{3}+b\frac{x^2}{2}+cx+C$

or $\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx+C$ , where C is any constant value.

Question 9: Find the integral intergration of $\int (2x^2+e^x)dx$

Answer:

Given intergral $\int (2x^2+e^x)dx$ ;

$\int 2x^{2}dx+\int e^{x}dx$

$=2\int x^{2}dx+\int e^{x}dx$

$=2\frac{x^3}{3}+e^x+C$

or $\frac{2x^3}{3}+e^x+C$ , where C is any constant value.

Question 10: Find the integral $\int {(\sqrt{x}-\frac{1}{\sqrt{x}})}^{2}dx$

Answer:

Given integral $\int {(\sqrt{x}-\frac{1}{\sqrt{x}})}^{2}dx$ ;

or $\int (x+\frac{1}{x}-2)dx$

$=\int xdx+\int \frac{1}{x}dx-2\int dx$

$=\frac{x^2}{2}+\ln |x|-2x+C$ , where C is any constant value.

Question 11: Find the integral intergration of $\int \frac{x^3+5x^2-4}{x^2}dx$

Answer:

Given intergral $\int \frac{x^3+5x^2-4}{x^2}dx$ ;

or $\int \frac{x^3}{x^2}dx+\int \frac{5x^2}{x^2}dx-4\int \frac{1}{x^2}dx$

$\int xdx+5\int 1dx-4\int x^{-2}dx$

$=\frac{x^2}{2}+5x-4\left(\frac{x^{-1}}{-1}\right)+C$

Or, $\frac{x^2}{2}+5x+\frac{4}{x}+C$ , where C is any constant value.

Question 12: Find the integral $\int \frac{x^3+3x+4}{\sqrt{x}}dx$

Answer:

Given intergral $\int \frac{x^3+3x+4}{\sqrt{x}}dx$ ;

or $\int \frac{x^3}{x^{\frac{1}{2}}}dx+\int \frac{3x}{x^{\frac{1}{2}}}dx+4\int \frac{1}{x^{\frac{1}{2}}}dx$

$=\int x^{\frac{5}{2}}dx+3\int x^{\frac{1}{2}}dx+4\int x^{-\frac{1}{2}}dx$

$=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left(x^{\frac{3}{2}}\right)}{\frac{3}{2}}+\frac{4\left(x^{\frac{1}{2}}\right)}{\frac{1}{2}}+C$

Or, $=\frac{2}{7}{x}^{\frac{7}{2}}+2x^{\frac{3}{2}}+8\sqrt{x}+C$ , where C is any constant value.

Question 13: Find the integral intergration of $\int \frac{x^3-x^2+x-1}{x-1}dx$

Answer:

Given integral $\int \frac{x^3-x^2+x-1}{x-1}dx$

It can be written as

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

Taking (x-1) common out

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

Now, cancelling out the term (x-1) from both numerator and denominator.

$=\int (x^2+1)dx$

Splitting the terms inside the brackets

$=\int x^{2}dx+\int 1dx$

$=\frac{x^3}{3}+x+c$

Question 14: Find the integral $\int (1-x)\sqrt{x}dx$

Answer:

Given intergral $\int (1-x)\sqrt{x}dx$ ;

$\int \sqrt{x}dx-\int x\sqrt{x}dx$ or

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

$=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C$

$=\frac{2}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}+C$ , where C is any constant value.

Question 15: Find the integral $\int \sqrt{x}(3x^2+2x+3)dx$

Answer:

Given intergral $\int \sqrt{x}(3x^2+2x+3)dx$ ;

$=\int 3x^2\sqrt{x}dx+\int 2x\sqrt{x}dx+\int 3\sqrt{x}dx$ or $=3\int x^{\frac{5}{2}}dx+2\int x^{\frac{3}{2}}dx+3\int x^{\frac{1}{2}}dx$

$=3\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+2\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+3\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C$

$=\frac{6}{7}{x}^{\frac{7}{2}}+\frac{4}{5}{x}^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$ , where C is any constant value.

Question 16: Find the integral $\int (2x-3\cos x+e^x)dx$

Answer:

Given integral $\int (2x-3\cos x+e^x)dx$ ;

splitting the integral as the sum of three integrals

$\int 2xdx-3\int \cos xdx+\int e^{x}dx$

$=2\frac{x^2}{2}-3\sin x+e^x+C$

$=x^2-3\sin x+e^x+C$ , where C is any constant value.

Question 17: Find the integral $\int (2x^2-3\sin x+5\sqrt{x})dx$

Answer:

Given integral $\int (2x^2-3\sin x+5\sqrt{x})dx$ ;

$2\int x^{2}dx-3\int \sin xdx+5\int \sqrt{x}dx$

$=2\frac{x^3}{3}-3(-\cos x)+5\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)+C$

$=\frac{2x^3}{3}+3\cos x+\frac{10}{3}{x}^{\frac{3}{2}}+C$ , where C is any constant value.

Question 19: Find the integral intergration of $\int \frac{se{c}^{2}x}{cose{c}^{2}x}dx$

Answer:

Given integral $\int \frac{se{c}^{2}x}{cose{c}^{2}x}dx$ ;

$\int \frac{\frac{1}{{\cos }^{2}x}}{\frac{1}{{\sin }^{2}x}}dx$

$=\int \frac{{\sin }^{2}x}{{\cos }^{2}x}dx$

$=\int ({\sec }^{2}x-1)dx$

$=\int {\sec }^{2}xdx-\int 1dx$

$=\tan x-x+C$ , where C is any constant value.

Question 20: Find the integral $\int \frac{2-3\sin x}{{\cos }^{2}x}dx$

Answer:

Given integral $\int \frac{2-3\sin x}{{\cos }^{2}x}dx$ ;

$\int (\frac{2}{{\cos }^{2}x}-\frac{3\sin x}{{\cos }^{2}x})dx$

Using antiderivative of trigonometric functions

$=2\tan x-3\sec x+C$ , where C is any constant value.

Question 21: Choose the correct answer
The anti derivative of $(\sqrt{x}+1/\sqrt{x})$ equals

$A)\frac{1}{3}{x}^{1/3}+2x^{1/2}+C$

$B)\frac{2}{3}{x}^{2/3}+\frac{1}{2}{x}^2+C$

$C)\frac{2}{3}{x}^{3/2}+2x^{1/2}+C$

$D)\frac{3}{2}{x}^{3/2}+\frac{1}{2}{x}^{1/2}+C$

Answer:

Given to find the anti derivative or integral of $(\sqrt{x}+1/\sqrt{x})$ ;

$\int (\sqrt{x}+1/\sqrt{x})dx$

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

$=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$

$=\frac{2}{3}{x}^{\frac{3}{2}}+2x^{\frac{1}{2}}+C$ , where C is any constant value.

Hence the correct answer is $\frac{2}{3}{x}^{3/2}+2x^{1/2}+C$.

Question 22: Choose the correct answer The anti derivative of

If $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$ such that f (2) = 0. Then f (x) is

$A)x^4+\frac{1}{x^3}-\frac{129}{8}$

$B)x^3+\frac{1}{x^4}-\frac{129}{8}$

$C)x^4+\frac{1}{x^3}+\frac{129}{8}$

$D)x^3+\frac{1}{x^4}-\frac{129}{8}$

Answer:

Given that the anti derivative of $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$

So, $\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}$

$f(x)=\int (4x^3-\frac{3}{x^4})dx$

$f(x)=4\int x^{3}dx-3\int x^{-4}dx$

$f(x)=4\left(\frac{x^4}{4}\right)-3\left(\frac{x^{-3}}{-3}\right)+C$

$f(x)=x^4+\frac{1}{x^3}+C$

Now, to find the constant C;

we will put the condition given, f (2) = 0

$f(2)=2^4+\frac{1}{2^3}+C=0$

$16+\frac{1}{8}+C=0$

or $C=\frac{-129}{8}$

$\Rightarrow f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}$

Therefore the correct answer is $x^4+\frac{1}{x^3}-\frac{129}{8}$