NCERT Solutions
Class 12 - Mathematics Part-2
Chapter 8: APPLICATION OF INTEGRALS

Complete NCERT Solutions for problems given in APPLICATION OF INTEGRALS chapter in Class 12 Mathematics Part-2.

Exercise 8.1

Question. 1

Find the area of the region bounded by the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).

Answer:

\(12\pi\)

Question. 2

Find the area of the region bounded by the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\).

Answer:

\(6\pi\)

Question.  3

Area lying in the first quadrant and bounded by the circle \(x^2 + y^2 = 4\) and the lines \(x = 0\) and \(x = 2\) is

(A)

\(\pi\)

(B)

\(\frac{\pi}{2}\)

(C)

\(\frac{\pi}{3}\)

(D)

\(\frac{\pi}{4}\)

Question.  4

Area of the region bounded by the curve \(y^2 = 4x\), y-axis and the line \(y = 3\) is

(A)

2

(B)

\(\frac{9}{4}\)

(C)

\(\frac{9}{3}\)

(D)

\(\frac{9}{2}\)

Miscellaneous Exercise on Chapter 8

Question. 1

Find the area under the given curves and given lines:

(i) \(y = x^2\), \(x = 1\), \(x = 2\) and x-axis

(ii) \(y = x^4\), \(x = 1\), \(x = 5\) and x-axis

Answer:

(i) \(\frac{7}{3}\)

(ii) \(624.8\)

Question. 2

Sketch the graph of \(y = |x + 3|\) and evaluate \(\int_{-6}^{0} |x + 3|\,dx\).

Answer:

\(9\)

Question. 3

Find the area bounded by the curve \(y = \sin x\) between \(x = 0\) and \(x = 2\pi\).

Answer:

\(4\)

Question.  4

Area bounded by the curve \(y = x^3\), the x-axis and the ordinates \(x = -2\) and \(x = 1\) is

(A)

\(-9\)

(B)

\(-\frac{15}{4}\)

(C)

\(\frac{15}{4}\)

(D)

\(\frac{17}{4}\)

Question.  5

The area bounded by the curve \(y = x|x|\), x-axis and the ordinates \(x = -1\) and \(x = 1\) is given by

(A)

0

(B)

\(\frac{1}{3}\)

(C)

\(\frac{2}{3}\)

(D)

\(\frac{4}{3}\)

NCERT Solutions Class 12 – Mathematics Part-2 – Chapter 8: APPLICATION OF INTEGRALS | Detailed Answers