JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

The area of the figure bounded by

y = \sin x,y = \cos x

in the first quadrant, is

View answer and explanation

Correct option: A

2\left( {\sqrt 2 - 1} \right)

Option A (

2\left( {\sqrt 2 - 1} \right)

) is correct because it represents the sum of the definite integrals of

(\cos x - \sin x)

from 0 to

\frac{\pi}{4}

and

(\sin x - \cos x)

from

\frac{\pi}{4}

\frac{\pi}{2}

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