Application of Integrals: Overview, Questions, Preparation

NCERT Maths 12th 2023 ( Ncert Solutions Maths class 12th )

Salviya Antony

Salviya AntonySenior Executive - Content

Updated on Jun 7, 2024 11:05 IST

NCERT Solutions Maths Class 12 Chapter 8 Application of Integrals are provided on this page. Application of Integrals Class 12 Solutions are really helpful for those who are preparing for the Class 12 CBSE board exams. Students can expect questions from this chapter for entrance exams also. Revising the Class 12 NCERT Solutions of Maths Chapter 8 Application of Integrals helps students to understand the concepts clearly.

The area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) is given by the formula, Area = ∫ab f(x) dx. Application of Integrals is a small chapter with only one exercise and one miscellaneous exercise. We have provided the accurate solutions of all the questions from this chapter. Check the NCERT Solutions Maths Class 12 Chapter 8 Application of Integrals PDF on this page.

NCERT Maths Class 12th Solution PDF - Application of Integrals Chapter Download

Check the NCERT Solutions Maths Class 12 Chapter 8 Application of Integrals below.

Download Here: NCERT Solution for Class XII Maths Application of Integrals PDF

Application of Integrals Solutions and FAQs

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

Q1. Sin 2x

A.1. d/dx cos 2x = -2 sin 2x

sin 2x = -1/2 d/dx cos 2x

Sin 2x = d/dx (1/2 cos 2x)

Therefore, an anti-derivative of sin 2x is -1/2 cos 2x

Q2.  cos 3x

A.2. d/dx sin 3x = 3 cos 3x

cos 3x = 1/3 dx sin 3x

cos 3x = d/dx (1/3 sin 3x)

Therefore, an anti-derivative of cos 3x is 1/3 sin 3x

Q3. e2x

A.3. d/dx (e2x)  = 2e2x

e2x = 1/2-d/dx(e2x)

e2x = d/dx(1/2 e2x)

Therefore, an anti-derivative of  e2x is 1/2 e2x

Q 1 2x / 1+x2

A 1 Let 1+x2 = t

2xdx = dt

log|t|= c

log |1+x2|+c

Q 2 log x2/x

A 2

Let log x = t

2xdx = dt

log|t|= c

log |1+x2|+c

Q 4 sin x sin (cos x)

A 4 put cos x = t

-sin x dx  = dt

sinx (sin (cosx dx)

Q 17 x /ex2

A 17 Put x= t

2xdx = dt

 

Q 22 sec^2 (7-4x)

A 22 Put 7-4x = t

-4dx = dt

-1/4 tan T+c

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