The area of the region bounded by the curve (y = x ) between the ordinates (x = 0 ), (x = 2 ) and the (x ) -axis is 2 sq units 4 sq units 3 sq units 1 sq units Correct Option (d)

We have (y = x,0 x 2 ) From the figure, area of the shaded region (A = 0 /2 xdx ) ( = [ - x] 0 /2 = - 2 + = 0 + 1 = 1 ) sq. units

class 12 maths application of integrals

The area of the region bounded by the curve $y = \sin x$ between the ordinates $x = 0$, $x = \frac{\pi }{2}$ and the $x$ -axis is
• 2 sq units
• 4 sq units
• 3 sq units
• 1 sq units
Correct Option (d)

VAVidaara Admin Asked 1mo ago 3 views 0 answers

📘 Application of Integrals NCERT Exemp. Ex. 1.3, Q. 30, Page 178 LA

The area of the region bounded by the curve $y = \sin x$ between the ordinates $x = 0$, $x = \frac{\pi }{2}$ and the $x$ -axis is

• 2 sq units

• 4 sq units

• 3 sq units

• 1 sq units

Correct Option (d)

VVidaara Team ✓ Verified solution NCERT & Exemplar

We have $y = \sin x,0 \le x \le \frac{\pi }{2}$

From the figure, area of the shaded region
$A = \int_0^{\pi /2} {\sin } xdx$

$= [ - \cos x]_0^{\pi /2} = - \cos \frac{\pi }{2} + \cos \theta = 0 + 1 = 1$ sq. units

No comments yet — start the discussion.