JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

The area in the first quadrant between

x^2+y^2=\pi^2

and

y=\sin x

View answer and explanation

Correct option: A

\frac{\left(\pi^3-8\right)}{4}

Option A (

\frac{\left(\pi^3-8\right)}{4}

) is correct because the area in the first quadrant between

x^2+y^2=\pi^2

and

y=\sin x

is the area of the quarter circle minus the area under the sine curve from 0 to

\pi

, which results in

\frac{\pi^3-8}{4}

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