Question

The area enclosed by the curves $y=\cos x, y=1+\sin 2 x$ and $x=\frac{3 \pi}{2}$ as $x$ varies from 0 to $\frac{3 \pi}{2}$, is

• A. $\frac{3 \pi}{2}-2$
• B. $\frac{3 \pi}{2}$
• C. $2+\frac{3 \pi}{2}$
• D. $1+\frac{3 \pi}{2}$

Answer 1

Answer: (C)

$ A=\int\limits_0^{3 \pi / 2}(1+\sin 2 x-\cos x) d x $
$=\left(x-\frac{1}{2} \cos 2 x-\sin x\right)_0^{\frac{3 \pi}{2}} $
$\left(\frac{3 \pi}{2}+\frac{1}{2}+1\right)-\left(0-\frac{1}{2}\right) $
$=2+\frac{3 \pi}{2} $