The area of the region bounded by the y-axis, y = c o s x and y = s i n x , 0 ≤ x ≤ π 2 is: (A) 2 sq. units (B) 2 +1sq. units (C) 2 − 1 sq. units (D) 2 2 − 1 sq. units

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t y − a x i s , y = c o s x , y = s i n x , 0 ≤ x ≤ π 2 Required area=∫0π4cosx dx−∫0π4sinx dx = [ s i n x ] 0 π 4 − [ − c o s x ] 0 π 4 = [ s i n π 4 − s i n 0 ] + [ c o s π 4 − c o s 0 ] = [ 1 2 − 0 + 1 2 − 1 ] = 2 2 − 1 = ( 2 − 1 ) s q . u n i t s H e n c e , t h e c o r r e c t o p t i o n i s ( c ) .

The area of the region bounded by the curve x 2 = 4 y and the straight line x = 4 y − 2 is: (A) 3 8 sq. units (B) 5 8 sq. units (C) 7 8 sq. units (D) 9 8 sq. units

This is an Objective Type Questions as classified in NCERT Exemplar G i v e n t h a t : T h e e q u a t i o n o f p a r a b o l a i s x 2 = 4 y … ( i ) a n d e q u a t i o n o f s t r a i g h t l i n e x = 4 y − 2 … ( i i ) S o l v i n g e q n ( i ) a n d ( i i ) w e g e t y = x 2 4 ⇒ x = 4 ( x 2 4 ) − 2 ⇒ x = x 2 − 2 ⇒ x 2 − x − 2 = 0 ⇒ x 2 − 2 x + x − 2 = 0 ⇒ x ( x − 2 ) + 1 ( x − 2 ) = 0 ⇒ ( x − 2 ) ( x + 1 ) = 0 ∴ x = − 1 , x = 2 Required area=∫−12x+24 dx−∫−12x24 dx = 1 4 [ x 2 2 + 2 x ] − 1 2 − 1 4 . 1 3 [ x 3 ] − 1 2 = 1 4 [ ( 4 2 + 2 ) − ( 1 2 − 2 ) ] − 1 1 2 [ 8 + 1 ] = 1 4 [ 6 + 3 2 ] − 1 1 2 [ 9 ] = 1 4 × 1 5 2 − 3 4 = 1 5 8 − 3 4 = 9 8 s q . u n i t s H e n c e , t h e c o r r e c t o p t i o n i s ( d ) .

The area of the region bounded by the curve y = 1 6 − x 2 and the x-axis is: (A) 8 sq. units (B) 2 0 π sq. units (C) 1 6 π sq. units (D) 2 5 6 π sq. units

This is an Objective Type Questions as classified in NCERT Exemplar H e r e , e q u a t i o n o f c u r v e i s y = 1 6 − x 2 Required area=2[∫0416−x2 dx] = 2 [ x 2 1 6 − x 2 + 1 6 2 s i n − 1 x 4 ] 0 4 = 2 [ ( 0 + 8 s i n − 1 4 4 ) − ( 0 + 0 ) ] = 2 [ 8 s i n − 1 ( 1 ) ] = 1 6 . π 2 = 8 π s q . u n i t s H e n c e , t h e c o r r e c t o p t i o n i s ( a ) .

Draw a rough sketch of the region { (x,y):y2≤6ax and x2+y2≤16a2 } . Also, find the area of the region sketched using the method of integration.

This is a long answer type question as classified in NCERT Exemplar W e a r e g i v e n t h a t { ( x , y ) : y 2 ≤ 6 a x a n d x 2 + y 2 ≤ 1 6 a 2 } E q u a t i o n o f P a r a b o l a i s y 2 = 6 a x … ( i ) a n d e q u a t i o n o f c i r c l e i s x 2 + y 2 = 1 6 a 2 … ( i i ) S o l v i n g e q n . ( i ) a n d ( i i ) w e g e t x 2 + 6 a x = 1 6 a 2 ⇒ x 2 + 6 a x − 1 6 a 2 = 0 ⇒ x 2 + 8 a x − 2 a x − 1 6 a 2 = 0 ⇒ x ( x + 8 a ) − 2 a ( x + 8 a ) = 0 ⇒ ( x + 8 a ) ( x − 2 a ) = 0 ∴ x = 2 a a n d x = − 8 a . ( Rejected as it out of region ) Area of the required shaded region=2[∫02a6axdx+∫2a4a16a2−x2] = 2 [ 6 a ∫ 0 2 a x d x + ∫ 2 a 4 a ( 4 a ) 2 − x 2 ] = 2 6 a . 2 3 [ x 3 / 2 ] 0 2 a + 2 [ x 2 ( 4 a ) 2 − x 2 + 1 6 a 2 2 s i n − 1 x 4 a ] 2 a 4 a = 4 6 3 a [ ( 2 a ) 3 / 2 − 0 ] + [ x ( 4 a ) 2 − x 2 + 1 6 a 2 s i n − 1 x 4 a ] 2 a 4 a = 4 6 3 a . 2 2 . a 3 / 2 + [ 0 + 1 6 a 2 s i n − 1 ( 4 a 4 a ) − 2 a 1 6 a 2 − 4 a 2 − 1 6 a 2 s i n − 1 2 a 4 a ] = 8 1 2 3 a 2 + [ 1 6 a 2 s i n − 1 ( 1 ) − 2 a 1 2 a 2 − 1 6 a 2 s i n − 1 1 2 ] = 1 6 3 3 a 2 + [ 1 6 a 2 . π 2 − 2 a . 2 3 a − 1 6 a 2 . π 6 ] = 1 6 3 3 a 2 + 8 π a 2 − 4 3 a 2 − 8 3 π a 2 = ( 1 6 3 3 − 4 3 ) a 2 + 1 6 3 π a 2 = 4 3 3 a 2 + 1 6 3 π a 2 = 4 3 ( 3 + 4 π ) a 2 s q . u n i t s H e n c e , t h e r e q u i r e d a r e a = 4 3 ( 3 + 4 π ) a 2 s q . u n i t s .

Maths NCERT Exemplar Solutions Class 12th Chapter Eight: Overview, Questions, Preparation

Updated on Jul 9, 2025 10:48 IST

By Payal Gupta, Retainer

Table of contents

Applications of Integrals Short Answer Type Questions
Applications of Integrals Long Answer Type Questions
Applications of Integrals Objective Type Questions

Applications of Integrals Short Answer Type Questions

1. Find the area of the region bounded by the curves $y^{2} = 9 x$ and $y = 3 x$ .

Sol:

\begin{array}{l} W e h a v e, y^{2} = 9 x, y = 3 x \\ S o l v i n g t h e t w o e q u a t i o n s, w e h a v e \\ {(3 x)}^{2} = 9 x \\ \Rightarrow 9 x^{2} - 9 x = 0 \Rightarrow 9 x (x - 1) = 0 \\ ∴ x = 0, 1 \\ A r e a o f t h e s h a d e d r e g i o n \\ = a r (r e g i o n O A B) - a r (Δ O A B) \\ = \int_{0}^{1} y_{1} d x = \int_{0}^{1} \sqrt[]{9 x} d x - \int_{0}^{1} 3 x d x \\ = 3 \int_{0}^{1} \sqrt[]{x} d x - 3 \int_{0}^{1} x d x = 3 \times \frac{2}{3} {[x^{3 / 2}]}_{0}^{1} - 3 {[\frac{x^{2}}{2}]}_{0}^{1} \\ = 2 [{(1)}^{3 / 2} - 0] - \frac{3}{2} [{(1)}^{2} - 0] = 2 (1) - \frac{3}{2} (1) = 2 - \frac{3}{2} = \frac{1}{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{1}{2} s q . u n i t s . \end{array}

2. Find the area of the region bounded by the parabolas $y^{2} = 2 p x$ and $x^{2} = 2 p y$ .

Sol:

$\begin{array}{l} W e a r e g i v e n t h a t x^{2} = 2 p y \dots (i) \\ a n d y^{2} = 2 p x \dots (i i) \\ F r o m e q n . (i) w e g e t y = \frac{x^{2}}{2 p} \\ P u t t i n g t h e t h e o f y i n e q n . (i i), w e h a v e \\ {(\frac{x^{2}}{2 p})}^{2} = 2 p x \Rightarrow \frac{x^{4}}{4 p^{2}} = 2 p x \\ \Rightarrow x^{4} = 8 p^{3} x \Rightarrow x^{4} - 8 p^{3} x = 0 \\ \Rightarrow x (x^{3} - 8 p^{3}) = 0 \\ ∴ x = 0, 2 p \\ Required a r e a = A r e a o f t h e r e g i o n (O C B A - O D B A) \\ = \int_{0}^{2 p} \sqrt[]{2 p x} d x - \int_{0}^{2 p} \frac{x^{2}}{2 p} d x \\ = \sqrt[]{2 p} \int_{0}^{2 p} \sqrt[]{x} d x - \frac{1}{2 p} \int_{0}^{2 p} x^{2} d x = \sqrt[]{2 p} . \frac{2}{3} {[x^{3 / 2}]}_{0}^{2 p} - \frac{1}{2 p} . \frac{1}{3} {[x^{3}]}_{0}^{2 p} \\ = \frac{2 \sqrt[]{2}}{3} [{(2 p)}^{3 / 2} - 0] - \frac{1}{6 p} [{(2 p)}^{3} - 0] \\ = \frac{2 \sqrt[]{2}}{3} \sqrt[]{p} . 2 \sqrt[]{2} p^{\frac{3}{2}} - \frac{1}{6 p} . 8 p^{3} \\ = \frac{8}{3} . p^{2} - \frac{8}{6} p^{2} = \frac{8}{6} p^{2} = \frac{4}{3} p^{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{4}{3} p^{2} s q . u n i t s . \end{array}$