JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

f(x)=\sin x \forall x \in\left[0, \frac{\pi}{2}\right], f(x)+f(\pi-x)=2 \quad \forall x \in\left[\frac{\pi}{2}, \pi\right]

and

f(x)=f(2 \pi-x) \forall x \in(\pi, 2 \pi]

. then the area enclosed by

y=f(x)

and

x

-axis is

View answer and explanation

Correct option: B

2 \pi

Option B is correct because the area is calculated by considering the symmetry and the given conditions for f(x).

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