Find the area of the region enclosed by the curve $y = x \sin x$ and the $x$-axis for $2 \pi \leq x \leq 3 \pi$.
A) $4 \pi$
B) 0
C) $5 \pi$
D) 4

Question

Quizplus · Accepted Answer

First, notice that the curve is non-negative and symmetric about the line $x=\pi$. Thus, we only need to find the area from $\pi$ to $3\pi$ and then double it. The area is given by:

\begin{align*}
\int_{2\pi}^{3\pi} x \sin x dx &= \left[ -x \cos x \right]_{2\pi}^{3\pi} - \int_{2\pi}^{3\pi} \cos x dx \
&= \left[ -x \cos x + \sin x \right]_{2\pi}^{3\pi} \
&= -3\pi(-1) + 2(-1) - [-2\pi(-1)+2(-1)] \
&= 5\pi
\end{align*}

Thus, the total area is $2(5\pi) = 10\pi$, which corresponds to choice C.

Find the Area of the Region Enclosed by the Curve y=xsin⁡xy = x \sin xy=xsinx

Find the Area of the Region Enclosed by the Curve $y = x \sin x$