Question

Consider the following statements
Statement I The area bounded by the curve $y=\sin x$ between $x=0$ and $x=2\pi$ is 2 sq units.
Statement II The area bounded by the curve $y=2\cos x$ and the $X$-axis from $x=0$ to $x=2\pi$ is 8 sq units.
Choose the correct option.

• A. Statement $I$ is true
• B. Statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (B)

I. We have, $y=\sin x$
Let us draw a graph of $\sin x$ between 0 to $2\pi$.

$\therefore$ Area of shaded region
$=\underset{0}{\overset{\pi }{\int }}\sin xdx+\left|\underset{\pi }{\overset{2\pi }{\int }}\sin xdx\right|$
$=[-\cos x{]}_0^{\pi }+\left|[-\cos x{]}_{\pi }^{2\pi }\right|$
$=[-\cos \pi +\cos 0]+|-\cos 2\pi +\cos \pi |$
$=|1+1|+|-1-1|$
$=2+2$
$=4$ sq units
II. We have, $y=2\cos x$
Let us draw the graph of $2\cos x$ between 0 to $2\pi$.

$\therefore$ Area of shaded region
$=\underset{0}{\overset{\pi /2}{\int }}2\cos xdx+\left|\underset{\pi /2}{\overset{3\pi /2}{\int }}2\cos xdx\right|+\underset{3\pi /2}{\overset{2\pi }{\int }}2\cos xdx$
$=2[\sin x{]}_0^{\pi /2}+\left|[2\sin x{]}_{\pi /2}^{3\pi /2}\right|+[2\sin x{]}_{3\pi /2}^{2\pi }$
$=2\left[\sin \frac{\pi }{2}-0\right]+\mid 2\left[\sin \frac{3\pi }{2}-\sin \frac{\pi }{2}\right]+2\left[\sin 2\pi -\sin \frac{3\pi }{2}\right]$
$=2+2\times 2+2$
$=2+4+2$
$=8$ sq units