Question

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

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

Answer 1

Answer: (D)

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