NCERT Solutions for Class 12 Maths Chapter 7

NCERT Solutions for Class 12 Maths – Free PDF Download

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Chapter 7 – Integrals 

NCERT Solutions for Class 12 Maths Chapter 7

7.1 Introduction

In Ch 7 Maths Class 12, the introduction part delves into the history of differential calculus and how integral calculus came into existence. You would learn how the original idea of derivative came about to solve the problem of defining tangent lines to the graphs of functions. It also deals with calculating the slope of these tangent lines.

This chapter will introduce the prime concept of the integral calculus, which involves trying to find the area bounded by the graph of these functions. You would learn how integration and differentiation are interrelated, and integration is the reverse process of differentiation. In differentiation, we are given a function, and we need to figure out the differentiation of that function. Conversely, in integration, you are given the differential of a function, and we need to find the function.

Other Key Concepts Discussed in This Chapter Are

• What are indefinite integrals and how they are denoted 

• What are definite integrals

• How the connection between the fundamentals of indefinite integrals and definite integrals serve as a practical tool for engineering and science. This connection is termed as the fundamental theorem of calculus.

With many different kinds of exercises and solved problems in this NCERT Solutions Class 12 Maths Chapter 7 you can easily revise all that you learn on this topic. 

7.2 Integration as a Reverse Process of Differentiation

In NCERT Solutions for Class 12 Chapter 7 Applications of Derivatives, you would learn how one can find the integral of a function given its derivative. The other name of integral is also anti-differentiation. You would understand terms like “arbitrary constant” which are varied to get different integrals of any given function. This “arbitrary constant” is also known as “constant of integration” and for any arbitrary real number K, ∫f(x) dx = F(x) + K. This is indefinite integral. You will get acquainted with a lot of basic formulas and properties of the indefinite integral in this section and how it is interpreted geometrically. 

This chapter would take you through a comparison between integration and differentiation and you will realize a few basics about these 2 parts of calculus like:

• Differentiation and integration are operations on functions

• They both satisfy linearity properties.

• There could be functions that are neither differentiable nor integrable.

• They differ in the sense differentiation of a function (wherever it exists) is a unique function while integral is not.

Exercise 7.1 Solutions: 22 Questions (21 Short Questions, 1 MCQ)

7.3 Methods of Integration

In this part of Integration NCERT Solutions, you will gain knowledge on other methods of finding integral apart from the method of inspection discussed in the above part. You would realize how this method is useful when an inspection is not suitable for many different kinds of functions. Here you will understand 3 different ways of integration i.e.:

• By Substitutions – This is used when a variable is substituted by another suitable variable it makes the integration process easier. So, if I = ∫f(x)dx and we put x = g(t) then we can write I = ∫ f[g(t)] g'(t) dt

• Using Partial Functions – If there are 2 polynomials p1(x) and p2(x) on x where p2(x) <>0 and degrees of p1(x) > p2(x) then we can say:

p1(x)/ p2(x) = t(x) + p’(x)/ p2(x). Here t(x) is a polynomial in x which is easily integrated and degree of p’(x) < degree of p2(x). 

• By Parts – If there are 2 functions f1(x) and f2(x) then we can say ∫[ f1(x)  f2(x)] dx = f1(x) ∫f2(x)] dx – ∫{f'(x) ∫f2(x)] dx} dx. This method is not applicable to the product of functions in all the cases like it will not work for ∫√x sin x dx.

You would further learn how to integrate using trigonometric identities for trigonometric functions with the use of many known trigonometric identities like:

• Sin2 Ө + Cos2 Ө = 1

• 2 Sin Ө Cos Ө = Sin 2 Ө

• 2 Sin Ө1. Cos Ө2 = Sin (Ө1 + Ө2) + Sin (Ө1 - Ө2)

Exercise 7.2 Solutions: 39 Questions (37 Short Questions, 2 MCQs)

Exercise 7.3 Solutions: 24 Questions (22 Short Questions, 2 MCQs)

7.4 Integrals of Some Particular Functions

In this part of NCERT Class 12 Maths Chapter 7 Solutions, you would learn many important integral formulas and how to apply them for many other standard integrals that are related to them. These short cut techniques provided in our solutions to problems on Integration Class 12 will help immensely in tackling many types of questions really fast.

You will also understand how to prove these standard integrals, some of those are mentioned below:

• ∫ dx/ (x2 – b2) = 1/2b log |x – b/x + a| + C

• ∫ dx/ (x2 + b2) = 1/b tan-1 x/b + C

• ∫ dx/ (√x2 + b2) = log |x + √x2 + b2| + C

Exercise 7.4 Solutions: 25 Questions (23 Short Questions, 2 MCQs)

7.5 Integration by Partial Functions

This unit of Chapter 7 Class 12 Maths NCERT solutions further expands what you learned in unit 7.3 about integration by partial functions. In this section, you would learn how an improper rational function can be converted into a proper rational function by a long division process. An improper rational function is the one where for 2 polynomials p1(x) and p2(x) on x,  p2(x) <>0 and degrees of p1(x) > p2(x) t.

You will learn here how to disintegrate an equation into parts and find the integral of each of the parts. This is known as partial fraction decomposition. 

Exercise 7.5 Solutions: 23 Questions (21 Short Questions, 2 MCQs)

7.6 Integration by Parts

This part would teach you methods on how to integrate products of functions. This method is also called partial integration. If for a single variable a, there are 2 differentiable functions f1 and f2, then by applying product rule of differentiation we can write d (f1f2)/da = f1df2/da + f2 df1/da.

You would then learn how to apply integration on both sides and conclude that integral of the product of 2 functions is given by:

(1st function) * (integral of the 2nd function) – Integral of [(differential coefficient of 1st function) * (integral of the 2nd function)] 

Exercise 7.6 Solutions: 24 Questions (22 Short Questions, 2 MCQ)

Exercise 7.7 Solutions: 11 Questions (9 Short Questions, 2 MCQ)

7.7 Definite Integral

In this section, you will closely learn about what a definite integral is. You would learn how to denote a definite integral and its notation; \[\int_{a}^{b}f(x)dx\] . In this, a is the lower limit and b is the upper limit of the integral.

You would also learn how to represent a definite integral either as a limit of a sum or an antiderivative in the interval [a, b]. 

Exercise 7.8 Solutions: 6 Questions (6 Short Questions)

7.8 Fundamental Theorem of Calculus

In this chapter, you would learn how concepts of integration and differentiation of a function are linked. This is done by calculating the anti-derivative difference at the upper and lower limits of the integration process. 

You Would Learn The Following

• Area function - \[\int_{a}^{b}f(x)dx\] defines the area of the curve y = f(x) which is bounded by ordinates of the x line (a and b). So, we can say that the area A(x) = \[\int_{a}^{b}f(x)dx\]

• First fundamental theorem of integral calculus – A’(x) = f(x) for all x ε [a,b], where f is a continuous function on the closed interval [a,b].

• 2nd fundamental theorem of integral calculus - \[\int_{a}^{b}f(x)dx\] = [F(x)]ba = F(b) – F(a), here f - is a continuous function on the closed interval [a,b]

          F – antiderivative of f.

Exercise 7.9 Solutions: 22 Questions (20 Short Questions, 2 MCQs)

7.9 Evaluation of Definite Integrals by Substitution

In this chapter, you would learn steps to evaluate definite integrals \[\int_{a}^{b}f(x)dx\] , by substitution as described below:

• Reduce the given integral to a known form by replacing y = f(x) and x = q(y) (considering the integral without limits).

• Now integrate the new integrand without mentioning the constant of integration.

Exercise 7.10 Solutions: 10 Questions (8 Short Questions, 2 MCQs)

7.10 Some Properties of Definite Integral

Here you will learn many properties of definite integral which will simplify the process of evaluating definite integrals. Some of the properties mentioned are as below:

• \[\int_{a}^{b}f(x)dx\] = \[\int_{a}^{b}f(t)dt\]

• \[\int_{0}^{2p}f(x)dx\] = \[\int_{0}^{p}f(x)dx\] + \[\int_{o}^{p}f(2p-x)dx\]

• \[\int_{0}^{p}f(x)dx\] = \[\int_{0}^{p}f(p-x)dx\]

• \[\int_{a}^{b}f(x)dx\] = \[\int_{a}^{p}f(x)dx\] + \[\int_{p}^{b}f(x)dx\]

You will also learn how to prove all these properties in this section of Integrals Class 12 Notes Maths Chapter 7.

Exercise 7.11 Solutions: 21 Questions (19 Short Questions, 2 MCQs)

Key Features of NCERT Solutions for Class 12 Maths Chapter 7

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• Solutions will help boost your confidence and check your preparedness for exams.