The area bounded by curve \(y=\sin 2x, x-\) axis and the lines \(x=\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) is
Answer & explanation
Correct answer: option 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.
$A=2\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin 2x\,dx$
$=2×\frac{1}{2}(1-0)=1$