(B) The area $A$ is given by the integral of the absolute value of the function because area cannot be negative.$A = \int_{-\pi/2}^{\pi/2} |\sin x| \,

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(B) The area $A$ is given by the integral of the absolute value of the function because area cannot be negative.$A = \int_{-\pi/2}^{\pi/2} |\sin x| \, dx$.Since $|\sin x|$ is an even function, we can write $A = 2 \int_{0}^{\pi/2} \sin x \, dx$.Evaluating the integral: $A = 2 [-\cos x]_{0}^{\pi/2}$.$A = 2 [-\cos(\pi/2) - (-\cos(0))]$.$A = 2 [0 - (-1)] = 2(1) = 2$ square units.

Class 12 Mathematics Application of Integration Area bounded by region of single curve

Medium

The area bounded by the curve $y = \sin x$ between $x = -\pi/2$ and $x = \pi/2$ is . . . . . . .

GUJCET 2025 All GUJCET

A
$4$
B
$2$
C
$3$
D
$1$

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Application of Integration

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Area bounded by region of single curve

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The area bounded by the curve $y = \sin x$ between $x = -\pi/2$ and $x = \pi/2$ is . . . . . . .

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