Question

Area bounded by the curves $y=x\sin x$ and $X-axis$ between $x=0$ and $x=2\pi$ is

• A. $2\pi$.sq.unit
• B. $3\pi$.sq.unit
• C. $4\pi$.sq.unit
• D. $5\pi$.sq.unit

Answer 1

The correct option is B $3\pi$.sq.unit

$\therefore$ Required Area$={\int }_0^{\pi }x\sin xdx+\left|{\int }_{\pi }^{2\pi }x\sin xdx\right|$
$={(-x\cos x+\sin x)}_0^{\pi }+\left|{(-x\cos x+\sin x)}_{\pi }^{2\pi }\right|$
$=(\pi +0)-(0+0)+|(-2\pi +0)-(\pi +0)|$
$=\pi +3\pi =4\pi$.sq.unit.