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MHTCET 15 Apr 2026 Shift 2 Paper

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Question : 131 of 150

Marks: +1, -0

\int_{0}^{\pi}|\sin 2 x| d x=

0
1
2
-1

Solution:

Concept:

The absolute value function must be split at the zeros of

\sin 2x

to remove the modulus.

Explanation:

The zeros of

\sin 2x

in

[0,\pi]

are

x=0

,

x=\frac{\pi}{2}

, and

x=\pi

.

So,

\sin 2x

is positive on

\left(0,\frac{\pi}{2}\right)

and negative on

\left(\frac{\pi}{2},\pi\right)

.

Hence,

|\sin 2x|=\sin 2x

for

0\le x\le \frac{\pi}{2}

, and

|\sin 2x|=-\sin 2x

for

\frac{\pi}{2}\le x\le \pi

.

Therefore,

\int_0^\pi |\sin 2x|\,dx=\int_0^{\pi/2}\sin 2x\,dx-\int_{\pi/2}^{\pi}\sin 2x\,dx

.

Since

\int \sin 2x\,dx=-\frac{\cos 2x}{2}

, we get:

\int_0^{\pi/2}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2}=1

.

\int_{\pi/2}^{\pi}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_{\pi/2}^{\pi}=-1

.

Thus,

\int_0^\pi |\sin 2x|\,dx=1-(-1)=2

.

Answer:

2

Option C.

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