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

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.1) Exercise 7.1

NCERT Solutions for class 12 Maths Chapter 7 Exercise 7.1 is designed by experienced teachers from Vedantu to improve your mathematical and analytical skills. The solutions are crafted in simple steps to give you a better understanding of the topic. Reference notes are also given for you to understand the basic concept of the topic. You can download the pdf format of the solutions from the official website of Vedantu. If you still have conceptual doubts that you can reach out to the teams of teachers available on the website. 

NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.1 starts with the introduction of Integral. This chapter covers the two types of Integrals i.e. Definite Integral and Indefinite Integral. The connection between these two types of integral is called the fundamental theorem of Calculus. The other topics under this chapter are integration as an inverse process of differentiation, geometrical interpretation of indefinite integral, properties of indefinite integral and comparison between differentiation and integration. Below are some brief excerpts of the theory that will help you in solving the anti derivatives and integrals in Exercise 7.1. 

Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)
Access NCERT Solutions for Class 12 Mathematics Chapter 7- Integrals part-1

Access NCERT Solutions for Class 12 Mathematics Chapter 7- Integrals

Exercise 7.1

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

sin $2x$

Ans:  We use the method of inspection as follows:

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

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

Thus, the anti-derivative of sin $2x$ is $-\frac{1}{2}\cos 2x$.

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

cos $3x$

Ans: We use the method of inspection as follows:

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

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

Thus, the anti - derivative of cos $3x$ is $\frac{1}{3}\sin 3x$.

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

${{e}^{2x}}$

Ans: We use the method of inspection as follows:

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

Thus, the anti-derivative of ${{e}^{2x}}$ is $\frac{1}{2}{{e}^{2x}}$.

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

${{\left( ax+b \right)}^{2}}$

Ans: We use the method of inspection as follows:

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

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

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

Thus, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\frac{1}{3a}{{\left( ax+b \right)}^{3}}$.

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

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

Ans: We use the method of inspection as follows:

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

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

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

Ans: 

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

$=4\int{{{e}^{3x}}dx}+\int{1}dx $ 

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

$=\frac{4}{3}{{e}^{3x}}+x+C $

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

Ans: 

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

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

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

8. $\int{\left( a{{x}^{2}}+bx+c \right)dx}$

Ans:

$\int{\left( a{{x}^{2}}+bx+c \right)}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+D $ 

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

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

Ans:  

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

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

Ans:

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

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

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

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

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

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

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

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

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

12. $\int{\frac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

Ans:$\int{\frac{{{x}^{3}}+3x+4}{\sqrt{x}}dx}$

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

$=\frac{2}{7}{{x}^{\frac{7}{2}}}+2{{x}^{\frac{3}{2}}}+8\sqrt{x}+C $

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

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

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

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

We obtain, on dividing:

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

$=\int{{{x}^{2}}dx}+\int{1.dx} $ 

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

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

Ans: $\int{\left( 1-x \right)}\sqrt{x}dx$

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

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

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

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

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

$=3\int{\left( {{x}^{\frac{5}{2}}}+2{{x}^{\frac{3}{2}}}+3{{x}^{\frac{1}{2}}} \right)} $  $=3\int{{{x}^{\frac{5}{2}}}dx+2\int{{{x}^{\frac{3}{2}}}dx+3\int{{{x}^{\frac{1}{2}}}}dx}} $  $=\frac{6}{7}{{x}^{\frac{7}{2}}}+\frac{4}{5}{{x}^{\frac{5}{2}}}+2{{x}^{\frac{3}{2}}}+C $

16. $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$

Ans: $\int{\left( 2x-3\cos x+{{e}^{x}} \right)dx}$          

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

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

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

17. $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

Ans: $\int{\left( 2{{x}^{2}}-3\sin x+5\sqrt{x} \right)}dx$

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

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

 $=\frac{2}{3}{{x}^{3}}+3\cos x+\frac{10}{3}{{x}^{\frac{3}{2}}}+C $

18. $\int{\sec x\left( \sec x+\tan x \right)dx}$

Ans: $\int{\sec x\left( \sec x+\tan x \right)dx}$

$=\int{\left( {{\sec }^{2}}x+\sec x\tan x \right)dx} $ 

$=\int{{{\sec }^{2}}xdx+\int{\sec x\tan xdx}} $ 

$=\tan x+\sec x+C $

19. $\int{\frac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

Ans: $\int{\frac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}dx}$

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

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

$=\int{{{\tan }^{2}}xdx} $ 

$=\int{{{\sec }^{2}}xdx}-\int{1dx} $ 

$=\tan x-x+C $

20. $\int{\frac{2-3\sin x}{{{\cos }^{2}}x}dx}$

Ans: $\int{\frac{2-3\sin x}{{{\cos }^{2}}x}dx}$

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

$=\int{2{{\sec }^{2}}xdx}-3\int{\tan x\sec xdx} $ 

$=2\tan x-3\sec x+C $ 

21. The anti – derivative of $\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)$ equals

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

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

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

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

Ans:

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

Thus, the correct answer is C.

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

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

  2. ${{x}^{3}}+\frac{1}{{{x}^{4}}}+\frac{129}{8}$

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

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

Ans: Given, $\frac{d}{dx}f\left( x \right)=4{{x}^{3}}-\frac{3}{{{x}^{4}}}$

  Anti-derivative of $4{{x}^{3}}-\frac{3}{{{x}^{4}}}=f\left( x \right)$

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

 $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 $ 

$Also, $ 

$f\left( 2 \right)=0 $ 

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

$\Rightarrow 16+\frac{1}{8}+C=0 $ 

$\Rightarrow C=\frac{-129}{8} $ 

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

 Thus, the correct answer is A.

NCERT Solutions For Exercise 7.1 Maths Class 12

Introduction

Calculus is a branch of Mathematics and it was developed by Sir Isaac Newton and G.W Leibnitz (a German mathematician) working independently in the latter half of the 17th century. Calculus can be broadly divided into two parts:

  1. Differential Calculus 

  2. Integral Calculus

Differential Calculus deals with the rate of change of some quantity like displacement, velocity, etc., which is also called finding the derivative of the quantity. 

Integral Calculus helps one to find the quantity when the rate of change of the quantity is known, which means that integration is the opposite process of differentiation. This is why integral of a function is also called anti-derivative of the function.

Integral Calculus Is Again Classified Into Two Types

  1. Indefinite Integral

  2. Definite Integral

Indefinite Integral is used to find a function if its derivatives is known and Definite Integral is used for finding the area bounded by a curve under certain conditions. 

Note: Earlier, the area of simple figures using straight lines like triangle, rectangle or trapezium, etc. was easy to calculate but finding the area under curved lines posed a problem. With the use of integral calculus this problem was solved. Use of Integral is extensive in various fields like Engineering, Physics, Economics, Biology, etc. 

Vedantu provides some techniques of integration that will help you solve the NCERT Solutions. Elementary properties of definite and indefinite integrals are also discussed below.

Integration As The Inverse Process Of Differentiation 

d/dx sin x= cosx , d/dx x2= 2x i.e. cos xis the derivation of sinx & 2x`is the derivative of  x2 . So we say that sinxis the integral of cosx and  x2 is the integral of 2x. 

In general, if d/dx {F(x) } =f(x), then we say that F(x) is the integral of f(x). This is why integral is called antiderivative.

If F(x) is the integral of f(x), we symbolically write ∫f(x) = F(x). Here the symbol ‘∫’ stands for integral.

Constant Of Integration

A function can have more than one integral as can be seen below:

As, d/dx sin x= cosx; d/dx (sin x + 1)= cosx;  d/dx (sin x + c) = cosx, where ‘c’ is a constant. 

∴∫cosx dx= sinx; ∫cosx dx = sinx + 1, ∫cosx dx = sinx + c. Thus sinx, sinx+1, sinx+c are all integrals of cosx

In this case, we can write ∫cosx dx= sinx+c where c is a constant. In general, is ∫f(x)dx = F(x),then we can also write ∫f(x)dx = F(x)+c where ‘c’ is a constant.

Geometrical Interpretation Of Indefinite Integral

The equation, ∫f(x)dx = F(x)+c, where ‘c’ is a constant and ‘c’ represents a family of curves and corresponds to different members of the family. These members can be derived by shifting any of the curves parallel to itself. This can be called a geometrical interpretation of Indefinite Integral.

Properties Of Indefinite Integrals

  1. Differentiation & integration operations are inverse of each other.
    if f(x) & F(x) are two functions such that F(x) = f(x) then d/dx {f(x)} = f(x)
    ∫f(x) dx = F(x) +c , where c is an arbitrary constant.

  2. Two indefinite integrals with the same derivative lead to the same family of curves. 

  3. If f(x) is a function having an antiderivative on an interval I , then k is a constant, then
    ∫k f(x) = k.∫ ∫(x)

  4. If f(x) & g(x) are two functions having antiderivative on an interval I, then the antiderivatives of f & g also exist on I and ∫{f(x) ± g(x)} dx= ∫f(x)dx ± ∫g(x)dx.

  5. The property No.4 can be extended to any number of functions.
    ∫{ f1(x) +k2, f2(x),........... kn, fn(x)} dx

  6. The properties of Np. 4 & 5 can be generalized in the following form
    ∫ [k1, f1(x) +k2, f2(x),........... kn, fn(x)] dx

∫ [k1, f1(x)dx +k2, f2(x)dx,........... kn, fn(x)dx] dx

Comparison Between Differentiation and Integration

  1. Both are operations on functions and the operations are inverse of each other.

  2. Both differentiation and integration operations satisfy the property of linearity.

  3. All functions may not be differentiable. Similarly all functions are not integral.

  4. When we differentiate a function, we get a unique function but when we integrate a function, the result may not be unique due to the presence of the constant term which may vary due to different conditions.

  5. When a polynomial is differentiated, we obtain another polynomial with a degree one less than the original polynomial. When a polynomial is integrated, we again get a polynomial but its degree is one more than the given polynomial.

  6. In the case of differentiation, we find the derivative at a point while in the case of integration, we integrate over an interval in which the integral is defined.

  7. The geometric interpretation of the derivative at a point i a curve is the slope of the tangent to the curve at that point. The indefinite integral of a function represents a family of curves, parallel to each other.

  8. As the derivative denotes the rate of change, it can be used to find the velocity of a moving particle when the displacement in time is known. Integration is used to find displacement in time when the velocity is known.

  9. Both differentiation and integration involve limits.

With the help of the above notes,you can now solve Exercise 7.1 of Chapter 7 Maths for Class 12, Integrals. Identifying the derivatives from the given expressions will be very easy for you. The solutions and the notes related to the subject provided by Vedantu will not only help the students understand the concept better but also solve the questions effectively. If you still have doubts then you can register with Vedantu and gain expertise in the subject.

Why Should You Download Ex 7.1 Class 12 Maths Solutions?

The reasons for downloading Class 12 Maths Chapter 7 Exercise 7.1 solution are mentioned below.

Clarify Doubts

There is no need to wait for the entire class to complete the first exercise. You can do it on your own and raise your doubts. Once you have jotted down your doubts, refer to the Ex 7.1 Class 12 Maths NCERT Solutions to find out how the experts have framed the answers. Compare yours with that of the experts to find out where you need to improve. In this way, you can easily clarify your doubts and move to the next level without any hassle.

Convenience At Its Best

There is no need to worry when you have the solutions for Class 12 Chapter 7 Maths solution with you. You can access it offline and refer to it whenever you want. Don’t waste your time waiting for the solution from your mentors. Grab the solution and understand how to approach integral problems from the very beginning.

Preparing A Base

As mentioned earlier, preparing a base in integration concepts is mandatory for the future. You will not only score well in the board exam but will also be able to crack the entrance exams. For this, you need to prepare your knowledge base using Exercise 7.1 Maths Class 12 Solutions.

Why Vedantu?

Vedantu is one of the most leading online education portals which gives a platform to the students to learn their subjects in a holistic environment. Our experienced teachers can teach you the shortcuts and tricks to solve the difficult questions. Vedantu also provides assistance for the IIT-JEE , KVPY and NEET examinations. Take the right decision to register with Vedantu and shape your career through us.

FAQs (Frequently Asked Questions)

1. What is the difference between definite and indefinite integrals?

Definite Integral is used for finding the area bounded by a curve under certain conditions whereas indefinite integral is used to find a function if its derivative is known.

2. What are the two types of integral calculus?

The two types of integral calculus are definite integral and indefinite integral.

3. What do you mean by differential calculus?

Differential Calculus deals with the rate of change of some quantity like displacement, velocity, etc., which is also called finding the derivative of the quantity.

4. How does NCERT Solutions provided by Vedantu help to prepare for examinations?

NCERT Solutions designed by Vedantu will definitely improve your analytical and mathematical skills. The solutions are given in step  wise as per the latest syllabus of CBSE board for you to get a better understanding of the solutions. There are enough practice worksheets which will help you with your revision. Solving the NCERT Solutions will boost your confidence level and thus prepare well for the exams.

Share this with your friends
SHARE
TWEET
SHARE
SUBSCRIBE