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

NCERT solutions for Class 12 Maths chapter 7 exercise 7.1 is the first exercise of chapter 7 Integrals. Basic concepts of integrals are discussed in this exercise. This exercise includes concepts pertaining to finding out integrals of basic functions like sinx,cosx etc. Exercise 7.1 Class 12 Maths can be a good source to grasp the initial concepts of integrals. Head Start with basic concepts is a must in Integrals to reach an advanced level which can be easily learnt from NCERT Solutions for Class 12 Maths chapter 7 exercise 7.1 provided below. Also it is a good source to score well in CBSE Cass 12 Board Exam. Basic integration problems are also asked in competitive exams like JEE Main. In subsequent exercises of Class 11 Maths NCERT book, students will cover advanced problems related to finding the area of a curve etc. The NCERT chapter Integrals also has the following exercise for practice.

  • Integrals Exercise 7.2

  • Integrals Exercise 7.3

  • Integrals Exercise 7.4

  • Integrals Exercise 7.5

  • Integrals Exercise 7.6

  • Integrals Exercise 7.7

  • Integrals Exercise 7.8

  • Integrals Exercise 7.9

  • Integrals Exercise 7.10

  • Integrals Exercise 7.11

  • Integrals Miscellaneous Exercise

Integrals Class 12 Chapter 7 Exercise: 7.1

Question:1 Find an anti derivative (or integral) of the following functions by the method of inspection. \sin 2x

Answer:

GIven \sin 2x ;

So, the anti derivative of \sin 2x is a function of x whose derivative is \sin 2x .

\frac{d}{dx}\left ( \cos 2x \right ) = -2\sin 2x

\implies \sin 2x =\frac{-1}{2} \frac{d}{dx}\left ( \cos 2x \right )

Therefore, we have \implies \sin 2x = \frac{d}{dx}\left ( \frac{-1}{2}\cos 2x \right )

Or, antiderivative of \sin 2x is \left ( \frac{-1}{2}\cos 2x \right ) .

Question:2 Find an anti derivative (or integral) of the following functions 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 following functions 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 following functions 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 following functions 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} (-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}) = \sin 2x -4e^{3x}

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

Question:6 Find the following integrals

\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 following integrals \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 following integrals \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 following integrals 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 following integrals \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 following integrals 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 following integrals \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 following integrals 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^2dx + \int 1dx

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

Question:14 Find the following integrals \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 following integrals \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 following integrals \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 following integrals \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:18 Find the following integrals \int \sec x ( \sec x + \tan x ) dx

Answer:

Given integral \int \sec x ( \sec x + \tan x ) dx ;

\int (\sec^2x+ \sec x \tan x ) \ dx

Using the integral of trigonometric functions

= \int (sec^2 x )\ dx+ \int \sec x \tan x\ dx

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

Question:19 Find the following integrals 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 following integrals \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} + 2 x ^{1/2}+ C \\\\ B) \frac{2}{3}x ^{2/3} + \frac{1}{2}x ^{2}+ C \\\\ C ) \frac{2}{3}x ^{3/2} + 2 x ^{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 option is (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) = 4 x ^3 - \frac{3}{x^4}

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

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

f(x) = 4\int x ^3 - 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 A .

More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.1

The NCERT Class 12 Maths chapter Integrals covers a total of 12 exercises including one miscellaneous exercise. Exercise 7.1 Class 12 Maths provides solutions to 22 main questions and their sub-questions. It includes basic questions related to finding integrals of basic functions. NCERT Solutions for Class 12 Maths chapter 7 exercise 7.1 must be referred to get a strong command on integrals.

Also Read| Integrals Class 12 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.1

  • The NCERT syllabus Class 12th Maths chapter 7 exercise provided here is in detail which is solved by subject matter experts .

  • Practicing exercise 7.1 Class 12 Maths can help students in a tremendous way to prepare for exams.

  • These Class 12 Maths chapter 7 exercise 7.1 solutions can be asked directly in the Board exams.

  • NCERT solutions for Class 12 Maths chapter 7 exercise 7.1 are highly recommended to students and can be used to solve physics questions also for the related concepts.

Also see-

  • NCERT Exemplar Solutions Class 12 Maths Chapter 7

  • NCERT Solutions for Class 12 Maths Chapter 7

NCERT Solutions Subject Wise

  • NCERT Solutions Class 12 Chemistry

  • NCERT Solutions for Class 12 Physics

  • NCERT Solutions for Class 12 Biology

  • NCERT Solutions for Class 12 Mathematics

Subject Wise NCERT Exemplar Solutions

  • NCERT Exemplar Class 12 Maths
  • NCERT Exemplar Class 12 Physics
  • NCERT Exemplar Class 12 Chemistry
  • NCERT Exemplar Class 12 Biology

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