NCERT Solutions For Class 12 Mathematics Chapter 8: Applications of the Integrals

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NCERT Solutions for class 12 mathematics Chapter 8 Applications of the Integrals cover important concepts of Area Between Two Curves, lines, parabolas; area of circles/ellipses. Application of Integrals covers the basic properties of integrals as well as the fundamental theorem of calculus. Applications of the Integrals will help students learn to find a function when its derivative is given and will also learn to find the area under a graph of a function.

Download: NCERT Solutions for Class 12 Mathematics Chapter 8 pdf

Class 12 Maths NCERT Solutions Chapter 8 Applications of the Integrals

Important Topics in Class 12 Mathematics Chapter 8 Applications of Integrals

Importabt concepts of Class 12 Maths covered in Chapter 8 Application Of Integrals of NCERT Solutions are:

Introduction to Applications of Integrals

The introduction section of this topic includes recollection of the idea of finding areas bounded by the curve. Definite integral as the limit of a sum, introduces different applications of integrals like the area under simple curves, between lines, parabolas and ellipses.

Average value of a function can be calculated using integration.

Example: Derivative of f(x) = x³ is f’(x) = 3x2; and the antiderivative of g(x) = 3x2 is f(x) = x³. Here, the integral of g(x) = 3x2 is f(x)=x³

Area Under Simple Curves

This section defines the area bounded by a curve. Area Under a Simple Curve is expressed using formula: y = f(x).

Area Between Two Curves

Area Between Two Curves section covers the method of finding the area between two curves with solved problems. Area can be found by dividing a certain region into a number of pieces of small area and then adding up the area of those tiny pieces. It is easier to find the area if the tiny pieces are vertical in shape.

Important Concepts of Area Between Two Curves:

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 = \(\oint_a^b y dx=\oint_b^a f(x) dx\)
Area of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b is given by the formula, Area = \(\oint_a^b\); where f(x) ≥ g(x) in [a, b]
If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], a < c < b, then Area = \(\oint_a^c + \oint_c^b\)

NCERT Solutions For Class 12 Maths Chapter 8 Exercises

The detailed solutions for all the NCERT Solutions for Chapter 8 Applications of Integrals under different exercises are as follows:

Exercise 8.1 Solutions: 13 Questions (8 Long, 3 Short, 2 MCQs)
Exercise 8.2 Solutions: 7 Questions(5 Long, 2 MCQs)
Miscellaneous Exercise Solutions: 19 Questions (8 Long, 7 Short, 4 MCQs)

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CBSE CLASS XII Related Questions

1.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:
- \(-\frac{\pi}{2}\)
- \(-\frac{\pi}{4}\)
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
View Solution
2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).
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3.
Find:

If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).
- \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
- \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
- \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
- \(p = 0, \, q = 0\)
View Solution
4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).
- \(0\)
- \(-2\)
- \(-1\)
- \(2\)
View Solution
5.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.
6.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]
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