JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

The area between the curve

y=\sin ^2 x, x

-axis and the ordinates

x=0

and

x=\frac{\pi}{2}

View answer and explanation

Correct option: B

\frac{\pi}{4}

Option B (

\frac{\pi}{4}

) is correct because the area under the curve

y = \sin^2 x

from 0 to

\frac{\pi}{2}

is found by integrating the function, which simplifies to

\frac{\pi}{4}

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