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 $\cos 3x$ .

$\frac{d}{dx}\left ( \sin 3x \right ) = 3\cos3x$

$\implies \cos 3x =\frac{1}{3} \frac{d}{dx}\left ( \sin 3x \right )$

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 $e ^{2x}$ .

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

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

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

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

Answer:

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

So, the anti derivative of

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

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

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

Answer:

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

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

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

$\int x^2 dx - \int1dx$

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

Question 8: Find the integral $\int ( ax ^2 + bx + c ) dx$

Answer:

Given intergral $\int ( ax ^2 + bx + c ) dx$ ;

$\int ax^2\ dx + \int bx\ dx + \int c\ dx$

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

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

or $\frac{ax^3}{3}+\frac{bx^2}{2}+cx +C$ , where C is any constant value.

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

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

$\int x\ dx + 5\int1 dx - 4\int x^{-2}\ dx$

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

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

$= \int x^2 \, dx + \int 1 \, dx$

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

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

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

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

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

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

Answer:

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

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

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

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

Question 19: Find the integral intergration of $\int \frac{sec ^2 x }{cosec ^2 x } dx$

Answer:

Given integral $\int \frac{sec ^2 x }{cosec ^2 x } dx$ ;

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

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

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

$=\int \sec^2 x\ dx-\int1 \ dx$

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

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

Answer:

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

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

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 $\left ( \sqrt x + 1/ \sqrt x \right )$ 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 $\left ( \sqrt x + 1/ \sqrt x \right )$ ;

$\int \left ( \sqrt x + 1/ \sqrt x \right )\ dx$

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

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

$= \frac{2}{3}x^{\frac{3}{2}} + 2x^{\frac{1}{2}} +C$ , 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 $\frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}$ 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, $\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$

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

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

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

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

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} $