Search any question & find its solution
Question:
Answered & Verified by Expert
Area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$ is
Options:
Solution:
1834 Upvotes
Verified Answer
The correct answer is:
$4$ sq. unit
We have $y=\sin x$
$\begin{array}{|c|c|c|c|c|c|c|} \hline x & 0 & \pi / 6 & \pi / 2 & \pi & 3 \pi / 2 & 2 \pi \\ \hline y & 0 & 0.5 & 1 & 0 & -1 & 0 \\ \hline\end{array}$
Join these points with a free hand to obtain a rough sketch
Required area $=($ area of $O A B)+($ area of $B C D)$
$=\int_0^\pi y d x+\int_\pi^{2 \pi}(-y) d x$
(Area $B C D$ is below $x$-axis)
$=\int_0^\pi \sin x d x-\int_\pi^{2 \pi} \sin x d x=4$ sq. unit.
$\begin{array}{|c|c|c|c|c|c|c|} \hline x & 0 & \pi / 6 & \pi / 2 & \pi & 3 \pi / 2 & 2 \pi \\ \hline y & 0 & 0.5 & 1 & 0 & -1 & 0 \\ \hline\end{array}$
Join these points with a free hand to obtain a rough sketch
Required area $=($ area of $O A B)+($ area of $B C D)$
$=\int_0^\pi y d x+\int_\pi^{2 \pi}(-y) d x$
(Area $B C D$ is below $x$-axis)
$=\int_0^\pi \sin x d x-\int_\pi^{2 \pi} \sin x d x=4$ sq. unit.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.