JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

Area under the curve

y=\sin 2 x+\cos 2 x

between

x=0

and

x=\frac{\pi}{4}

, is

View answer and explanation

Correct option: B

1 sq. unit

Option B (1 sq. unit) is correct because the area is calculated by integrating

\sin 2x + \cos 2x

from

x=0

x=\frac{{\pi}}{{4}}

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