Question

The area (in sq. units) bounded between $y=3sinx$ and $y=-4si{n}^{3}x$ from $x=0$ to $x=\pi$ is

• A. $4\pi$
• B. $34\pi$
• C. $4$
• D. $\frac{34}{3}$

Answer 1

Answer: (D)

As $3\sin x\ge 0$ and $-4{\sin }^{3}x\le 0\forall x\in [0,\pi ]$,
the required area is $A={\int }_0^{\pi }\left(3\sin x-\left(-4{\sin }^{3}x\right)\right)dx$
$A={\int }_0^{\pi }(3\sin x+(3\sin x-\sin 3x))dx$
$={\int }_0^{\pi }(6\sin x-\sin (3x))dx$
$={\left[-6\cos x+\frac{\cos (3x)}{3}\right]}_0^{\pi }$
$=\left(6-\frac{1}{3}\right)-\left(-6+\frac{1}{3}\right)$
$=12-\frac{2}{3}=\frac{34}{3}$ sq. units