The area bounded by curve \(y=\sin 2x, x-\) axis and the lines \(x=\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) is

\(1\)

The correct answer is Option (1) → 1

The required area (A) is,

$A=\int\limits_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin 2x\,dx$

$=-\frac{1}{2}\left[\cos 2x\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$

$=0$

Since, the function $y=\sin 2x$ is symmetric about its midpoint in the given interval, the positive and negative areas cancels out.

$A=2\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin 2x\,dx$

$=2×\frac{1}{2}(1-0)=1$