# RD Sharma Class 12 Chapter 20 Solutions (Definite Integrals)

RD Sharma Class 12 Maths Solutions Chapter 20 Definite Integrals discusses the topics of the chapter with you in detail. A thorough understanding of these topics is essential in order to appear in CBSE and engineering entrance exams such as JEE Mains and JEE Advanced. The topics you’ll learn in this chapter of RD Sharma Class 12 Maths Solutions are Fundamental theorem of Integral Calculus, evaluation of definite integrals, evaluating definite integrals by substitution, integration as the limit of a sum, and properties of definite integrals.

This chapter consists of 4 exercises and a total of 193 questions explaining the concepts in detail. On practising these questions you’ll be able to answer highly complex problems on Definite Integrals. The book has been designed keeping your CBSE and competitive exams in consideration. All the solutions are curated so as to make the concept clear and fun.

Instasolv’s maths experts provide solutions to these exercises in the easiest possible way. They keep complexities low and help you to practise Definite Integrals and other concepts that have been implemented here. The solutions are available in the same arrangement as you find them in RD Sharma Solutions for Chapter 20 Definite Integrals.

## Topics discussed in RD Sharma Class 12 Maths Solutions Chapter 20 Definite Integrals

**Fundamental theorem of Integral Calculus**

Consider F(x) the antiderivative of the primitive of a continuous function f(x). F is defined on the closed interval [a,b] which means d/dx(F(x) = f(x).

Then f(x)’s definite integral over the closed interval [a,b] can be denoted by

Call a and b as the integration limits where a is the lower limit and b is the upper limit.

Consider the closed interval [a,b] the integration interval.

**Evaluation of Definite Integrals**

In order to evaluate the definite integral of f(x), the continuous function that is defined on the closed integration interval [a,b], we can use the following algorithm.

- Evaluate the indefinite integral ∫f(x)dx. Call this F(x). You may skip the integration constant.
- Now evaluate F(b) and F(a)
- Now calculate F(b) – F(a)

**Evaluating Definite Integrals by Substitution**

Consider g(x) = t

therefore, g’(x) dx = dt.

And when x = a and t = g(a)

t = g(b) when x = b.

Therefore,

is equal to

**Properties of Definite Integrals**

- p∫q f(a) da = p∫q f(t) dt
- p∫q f(a) d(a) = – q∫p f(a) d(a), Also p∫p f(a) d(a) = 0
- p∫q f(a) d(a) = p∫r f(a) d(a) + r∫q f(a) d(a)
- p∫q f(a) d(a) = p∫q f( p + q – a) d(a)
- o∫p f(a) d(a) = o∫p f(p – a) d(a)
- ∫02p f(a)da = ∫0p f(a)da +∫0p f(2p-a)da…if f(2p-a) = f(a)
- ∫02 f(a)da = 2∫0a f(a) da … if f(2p-a) = f(a)
- ∫02 p f(a)da = 0 … if f(2p-a) = -f(a)
- ∫-pp f(a)da = 2∫0p f(a) da … if f(-a) = f(a) or it’s an even function
- ∫-ppf(a)da = 0 … if f(2p-a) = -f(a) or it’s an odd function

**Integration as the limit of a sum**

Use Definite Integrals when you know that limits can be defined or should be defined. This is usually done when you are looking for a unique value.

Let f(x) = Continuous real-valued function

F(x) is defined on the closed interval [a,b] further divided into s equal parts each with a width ‘q’ by adding (s-1) points a+q, a+2q, a+3q, a+4q,…., a+(s-1)q between the two points a and b.

Then sq = b-a

Or q = (b-a)/s

### Discussion of exercises in RD Sharma Class 12 Maths Solutions Chapter 20 Definite Integrals

- The first exercise of Definite Integrals contains 59 questions and talks about the method of evaluating definite integrals.
- The second exercise of Definite Integrals i.e. 20. 2 will test your understanding of the topic Evaluation of Definite Integrals by Substitution in its 53 questions questionnaire.
- This exercise 20.3 talks about different properties of Definite Integral in 49 questions.
- Exercise 20.4 discusses 32 questions on integration as the limit of a sum.

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