Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
0 votes

Find the area bounded by the curve \(y = \sin\: x\) between \(x = 0\) and \(x = 2π\).

$\begin{array}{1 1} 4 \\ 2 \\ 6 \\ 8 \end{array}$

Can you answer this question?

1 Answer

0 votes
The required area is
A=area of OAB+area of BCD
$\;\;=\int_0^\pi ydx+\int_\pi^{2\pi}-ydx$
The curve between the limits 0 to $\pi$ is on the positive side of y-axis
The curve between the limits $\pi$ to $2\pi$ is on the negative side of y-axis
Hence $A=\int_0^\pi\sin xdx+\int_\pi^{2\pi}\sin xdx$
$\;\;\;\;\;\;\;\;\;A=[-\cos x]_0^\pi+[-\cos x]_\pi^{2\pi}$
Hence the required area is 4 sq,units.
answered Dec 21, 2013 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App