Class 12 maths NCERT Solutions Exercise 7.1-Integral

Class 12 maths NCERT Solutions Exercise 7.1 of Chapter 7-Integrals-Here NCERT solutions of class 12 maths exercise 7.1 of chapter 12 are the solutions of the chapter 7 exercise 7.1-integrals of class maths NCERT text book . All these NCERT solutions are well explained by the maths expert through a step by step method so that every student could be assessable to the exercise 7.1 of chapter 7-Integral. The purpose of these NCERT Solutions of exercise 7.1 is to clear the concept of integration among the students of class 12.

Class 12 maths NCERT Solutions Exercise 7.1-Integral

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What is Integrals? Integration is just opposite to differentiation,to understand integration let’s have a function (3x +2), its derivative is 3.Then we can say integral of 3 is 3x +C where C =2 is known as an arbitrary constant. It can be cleared by the following examples.

y = 3x, dy/dx=3

y =3x +2,dy/dx =3

y= 3x +3,dy/dx =3

The differentiation of y with respect to x is the same 3, so we can say that integration of 3 is 3x +C, where C may be any number, further, we can say that 3x +C is the indefinite integration(integral) of 3 , in other words, it can also be called that 3x+C is the anti-derivative of 3.

If there is a function f(x) such that its derivative is g(x),then f(x) is anti-derivative(Integral) of the function g(x).Therefore when the differentiation of a function is integrated we get the original function,in simple words integration is opposite of diffrentiation.

The d/dx(xⁿ) = nx^n-1

Here, integral of nx^n-1is xⁿ

Integration is represented by the sign ∫ and dx is attached at the end of the function which means integration of the function with respect to x. The integral of a function is calculated as follows.

$\boldsymbol{\int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$

Application of integration: When differentiated parts of a region within limits combine together we get the integral of the function within the region. So, integration is used to calculate physical quantities like area, volume, mass, etc.It is widely used in science and technology for the evaluation of the different type of physical quantities and in business predicting the outcomes of particular inputs applied.

Class 12 Maths NCERT solutions of Chapter 7 Integrals

Exercise 7.1- Integrals

Exercise 7.2-Integral

Exercise 7.3-Integrals

Exercise 7.4 -Integral

NCERT Solutions Exercise 7.1-Integral

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

Sin 2x

Ans.

We know -sinθ is the derivative of cosθ,so differentiation of cosθ is =-sin

$\boldsymbol{\frac{d}{dx}\left ( cos2x \right )= -2sin2x}$

$\boldsymbol{\Rightarrow sin2x=-\frac{1}{2}\frac{d}{dx}\left ( cos2x \right )}$

$\boldsymbol{\Rightarrow sin2x=\frac{d}{dx}\left (-\frac{1}{2} cos2x \right )}$

Hence anti-derivative of sin2x is -1/2(cos2x).

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

cos 3x

Ans. We know cosθ is the derivative of sinθ

$\boldsymbol{\frac{d}{dx}\left ( cos3x \right )}$

$\boldsymbol{\frac{d}{dx}\left ( sin3x \right )=3cosx}$

$\boldsymbol{\Rightarrow cos3x=\frac{1}{3}\frac{d}{dx}\left ( sin3x \right )}$

$\boldsymbol{\Rightarrow cos3x=\frac{d}{dx}\left ( \frac{1}{3}sin3x \right )}$

Therefore anti-derivative (integral) of cos3x is 1/3(sin3x).

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Q3.Find an anti-derivative (or integral) of the following functions by the method of inspection

$\boldsymbol{e^{2x}}$

Ans.We know the derivative of $\boldsymbol{e^{x}}$ is $\boldsymbol{e^{x}}$ ,so

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

$\boldsymbol{\Rightarrow e^{2x}=\frac{1}{2}\frac{d}{dx}\left ( e^{2x} \right )}$

$\boldsymbol{\Rightarrow e^{2x}=\frac{d}{dx}\left (\frac{1}{2} e^{2x} \right )}$

Therefore anti-derivative of $\boldsymbol{e^{2x}}$ is 1/2( $\boldsymbol{e^{2x}}$ ).

Q4.Find an anti-derivative (or integral) of the following functions by the method of inspection

(ax +b)²

Ans.We know the derivative of x³ is 3x², so

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

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

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

Therefore antiderivative of (ax+b)² is 1/3a(ax+b)³

Class 12 Maths NCERT solutions of Chapter 7 Integrals

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

$\boldsymbol{sin2x-4e^{3x}}$

Ans.We know -sinθ is the derivative of cosθ

$\boldsymbol{\frac{d}{dx}\left ( cos2x \right )=-2sin2x}$

$\boldsymbol{\Rightarrow sin2x=-\frac{1}{2}\frac{d}{dx}\left ( cos2x \right )}$

$\boldsymbol{\Rightarrow sin2x=\frac{d}{dx}\left ( -\frac{1}{2}cos2x \right )...\left ( i \right )}$

We know derivative of $\boldsymbol{e^{x}}$ is $\boldsymbol{e^{x}}$

$\boldsymbol{\frac{d}{dx}\left ( e^{3x} \right )=3e^{3x}}$

$\boldsymbol{e^{3x}=\frac{1}{3}\frac{d}{dx}\left ( e^{3x} \right )}$

$\boldsymbol{\Rightarrow 4e^{3x}=\frac{4}{3}\frac{d}{dx}\left ( e^{3x} \right )}$

$\boldsymbol{\Rightarrow 4e^{3x}=\frac{d}{dx}\left ( \frac{4}{3}e^{3x} \right )}$

$\boldsymbol{\Rightarrow 4e^{3x}=\frac{d}{dx}\left ( \frac{4}{3}e^{3x} \right )...\left ( ii \right )}$

Subtracting equation (ii) from (i), we get

$\boldsymbol{sin2x-4e^{x}=\frac{d}{dx}\left ( -\frac{1}{2}cos2x-\frac{4}{3}e^{3x} \right )}$

Therefore antiderivative of $\boldsymbol{sin2x-4e^{3x}}$ is $\boldsymbol{\left ( -\frac{1}{2}cos2x-\frac{4}{3}e^{3x} \right )}$

Find the following integrals from Exercises 6 to 20.

Q6. $\boldsymbol{\int \left ( 4e^{3x} +1\right )dx}$

Ans.

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

$\boldsymbol{\Rightarrow 4\int e^{3x}dx+\int 1dx}$

$\boldsymbol{\Rightarrow 4.\frac{e^{3x}}{3}+x+C}$

$\boldsymbol{Q7.\: \int x^{2}\left ( 1-\frac{1}{x^{2}} \right )dx}$

Ans.

$\boldsymbol{ \int x^{2}\left ( 1-\frac{1}{x^{2}} \right )dx}$

$\boldsymbol{\Rightarrow \int \left (x^{2}-1 \right )dx}$

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

$\boldsymbol{\Rightarrow \frac{x^{3}}{3}-x+C}$

$\boldsymbol{Q8.\: \int \left ( ax^{2}+bx+c \right )dx}$

Ans.

$\boldsymbol{ \int \left ( ax^{2}+bx+c \right )dx}$

$\boldsymbol{\Rightarrow a\int x^{2}dx+b\int xdx+c\int 1 dx}$

$\boldsymbol{\Rightarrow a.\frac{x^{3}}{3}+b.\frac{x^{2}}{2}+cx+C}$

Class 12 Maths NCERT solutions of Chapter 7 Integrals

$\boldsymbol{Q9.\, \int \left ( 2x^{2}+e^{x} \right )dx}$

Ans.

$\boldsymbol{\int \left ( 2x^{2}+e^{x} \right )dx}$

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

$\boldsymbol{\Rightarrow 2.\frac{x^{3}}{3}+e^{x}+C}$

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

$\boldsymbol{Q10.\: \int \left ( \sqrt{x}-\frac{1}{\sqrt{x}} \right )^{2}dx}$

Ans.

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

Writting the function in expanded form

$\boldsymbol{ \int \left ( x+\frac{1}{x} -1\right )dx}$

$\boldsymbol{ \Rightarrow \int xdx+\int \frac{1}{x}dx -\int 1dx}$

$\boldsymbol{\Rightarrow \frac{x^{2}}{2}+log\left | x \right |-x+C}$

$\boldsymbol{Q11.\: \int \frac{x^{3}+5x^{2}-4}{x^{2}}dx}$

Ans.

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

$\boldsymbol{\Rightarrow \int \left ( x+5-4x^{-2} \right )dx}$

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

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

$\boldsymbol{\Rightarrow \frac{x^{2}}{2}+5x+\frac{4}{x}+C}$

$\boldsymbol{Q12.\int \left ( \frac{x^{3}+3x+4}{\sqrt{x}} \right )dx}$

Ans.

$\boldsymbol{\int \left ( \frac{x^{3}+3x+4}{\sqrt{x}} \right )dx}$

$\boldsymbol{\Rightarrow \int \left ( x^{\frac{5}{2}} +3x^{\frac{1}{2}}+4x^{-\frac{1}{2}}\right )dx}$

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

$\boldsymbol{\Rightarrow \frac{x^{7/2}}{7/2}+3.\frac{x^{3/2}}{3/2}+4.\frac{x^{1/2}}{1/2}+C}$

$\boldsymbol{\Rightarrow \frac{2}{7}.x^{7/2}+2x^{3/2}+8x^{1/2}+C}$

$\boldsymbol{\Rightarrow \frac{2}{7}.x^{7/2}+2x^{3/2}+8\sqrt{x}+C}$

Class 12 Maths NCERT solutions of Chapter 7 Integrals

$\boldsymbol{Q13.\int \frac{x^{3}-x^{2}+x-1}{x-1}dx}$

Ans.

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

Dividing numerator by the denominator, we get x² +1

$\boldsymbol{\int \left ( x^{2}+1 \right )dx}$

$\boldsymbol{\Rightarrow \int \ x^{2}dx+\int 1 dx}$

$\boldsymbol{\Rightarrow\frac{x^{3}}{3}+x+C}$

$\boldsymbol{Q14.\int \left ( 1-x \right )\sqrt{x}.dx}$

Ans.

$\boldsymbol{\int \left ( 1-x \right )\sqrt{x}.dx}$

$\boldsymbol{=\int \left ( \sqrt{x}-x^{3/2} \right )dx}$

$\boldsymbol{=\int x^{1/2}dx-\int x^{3/2} dx}$ + C

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

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

$\boldsymbol{Q15.\int \sqrt{x}\left ( 3x^{2} +2x+3\right )dx}$

Ans.

$\boldsymbol{\int \sqrt{x}\left ( 3x^{2} +2x+3\right )dx}$

$\boldsymbol{\Rightarrow 3\int x^{5/2}dx+2\int x^{3/2}dx+3\int x^{1/2}dx}$

$\boldsymbol{\Rightarrow 3.\frac{x^{7/2}}{7/2}+2.\frac{x^{5/2}}{5/2}+3.\frac{x^{3/2}}{3/2}}$ + C

$\boldsymbol{\Rightarrow \frac{6}{7}x^{7/2}+\frac{4}{5}.x^{5/2}+2x^{3/2}+C}$

$\boldsymbol{Q16.\: \int \left ( 2x-3cosx+e^{x} \right )dx}$

Ans.

$\boldsymbol{ \int \left ( 2x-3cosx+e^{x} \right )dx}$

$\boldsymbol{\Rightarrow 2\int x\: dx-3\int cosx\: dx+\int e^{x} dx}$

$\boldsymbol{\Rightarrow 2.\frac{x^{2}}{2}+3sinx+e^{x}+C}$

$\boldsymbol{\Rightarrow x^{2}-3sinx+e^{x}+C}$

$\boldsymbol{Q17.\int \left ( 2x^{2} -3sinx+5\sqrt{x}\right )dx}$

Ans.

$\boldsymbol{\int \left ( 2x^{2} -3sinx+5\sqrt{x}\right )dx }$

$\boldsymbol{\Rightarrow 2\int x^{2} -3\int sinx\: dx+5\int \sqrt{x}dx }$

$\boldsymbol{\Rightarrow 2\int x^{2}dx -3\int sinx\: dx+5\int x^{1/2}dx }$

$\boldsymbol{\Rightarrow 2.\frac{x^{3}}{3}+3cosx+5.\frac{x^{3/2}}{3/2} +C}$

$\boldsymbol{\Rightarrow \frac{2}{3}x^{3}+3cosx+\frac{10}{3}x^{3/2}+C}$

$\mathbf{Q18.\int secx(secx + tanx)dx}$

Ans. On multiplication, we get,

$=\mathbf.\int (sec^{2}x + secx\times tanx)dx$

By taking separately, we get,

$=\mathbf.\int sec^{2}\, xdx \, +\int secx\, tan\, xdx$

We get,

= tan x + sec x + C

$Q19.\int \frac{sec^{2}x}{cosec^{2}x}dx$

Ans. We get,

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

So,

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

We get,

$=\int tan^{2}xdx$

On further calculation, we get,

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

By taking seriously, we get

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

= tan x – x + C

Class 12 Maths NCERT solutions of Chapter 7 Integrals

$\boldsymbol{Q20.\int \frac{2-3sinx}{cos^{2}x}dx}$

Ans.

$\boldsymbol{\int \frac{2-3sinx}{cos^{2}x}}$

$\boldsymbol{=\int \frac{2}{cos^{2}x}dx-\int \frac{3sinx}{cos^{2}x}dx}$

$\boldsymbol{=2\int sec^{2}xdx-3\int tanx.secxdx}$

$\boldsymbol{=2tanx-3secx+C}$

$\boldsymbol{Q21.The\: anti-derivative \: of \: \left ( \sqrt{x} +\frac{1}{\sqrt{x}}\right )\: is}$

$\boldsymbol{\left ( A \right )\frac{1}{3}x^{1/3}+2x^{1/2}+C}$

$\boldsymbol{\left ( B\right )\frac{2}{3}x^{2/3}+\frac{1}{2}x^{2}+C}$

$\boldsymbol{\left ( C\right )\frac{2}{3}x^{3/2}+2x^{1/2}+C}$

$\boldsymbol{\left ( D\right )\frac{3}{2}x^{3/2}+\frac{1}{2}x^{1/2}+C}$

Ans.

$\boldsymbol{\int \left ( \sqrt{x} +\frac{1}{\sqrt{x}}\right )dx}$

$\boldsymbol{\Rightarrow \int x^{1/2}dx +\int x^{-1/2}dx}$

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

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

Therefore the correct answer is (C)

$\boldsymbol{Q22.If \: \frac{d}{dx}f\left ( x \right )=4x^{3}-\frac{3}{x^{4}}\: such\: that\: f\left ( 2 \right )=0\, then \: f\left ( x \right )\: is}$

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

$\boldsymbol{\left ( B \right )x^{3}+\frac{1}{x^{4}}+\frac{129}{8}}$

$\boldsymbol{\left ( C \right )x^{4}+\frac{1}{x^{3}}+\frac{129}{8}}$

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

Ans. We are given

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

Therefore anti-derivative of $\boldsymbol{4x^{3}-\frac{3}{x^{4}}}$ is f(x)

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

$\boldsymbol{\Rightarrow 4\int x^{3}dx-\int \frac{3}{x^{4}}dx}$

$\boldsymbol{\Rightarrow4 \int x^{3}-3\int x^{-4}}$

$\boldsymbol{\Rightarrow4\frac{x^{4}}{4}-3\frac{x^{-3}}{-3}+C}$

$\boldsymbol{\Rightarrow x^{4}+\frac{1}{x^{3}}+C}$

We are given f(2) =0

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

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

$\boldsymbol{C+\frac{129}{8}=0\Rightarrow C= -\frac{129}{8}}$

Substituting this value of C in f(x)

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

Therefore the correct answer is (A)

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Class 12 maths NCERT Solutions Exercise 7.1-Integral

Class 12 maths NCERT Solutions Exercise 7.1-Integral

Class 12 Maths NCERT solutions of Chapter 7 Integrals

NCERT Solutions Exercise 7.1-Integral

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Class 12 Maths NCERT solutions of Chapter 7 Integrals

Class 12 Maths NCERT solutions of Chapter 7 Integrals

Class 12 Maths NCERT solutions of Chapter 7 Integrals

Class 12 Maths NCERT solutions of Chapter 7 Integrals

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Class 12 maths NCERT Solutions Exercise 7.1-Integral

Class 12 maths NCERT Solutions Exercise 7.1-Integral

Class 12 Maths NCERT solutions of Chapter 7 Integrals

NCERT Solutions Exercise 7.1-Integral

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Class 12 Maths NCERT solutions of Chapter 7 Integrals

Class 12 Maths NCERT solutions of Chapter 7 Integrals

Class 12 Maths NCERT solutions of Chapter 7 Integrals

Class 12 Maths NCERT solutions of Chapter 7 Integrals

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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NCERT Solutions for class 9 science

NCERT Solutions for class 10 maths

NCERT Solutions for Class 10 Science

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