Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

The area of the region bounded by the graph of $y=\sin x$ and $y=\cos x $ between $x=0$ and $x=\large\frac{\pi}{4}$ is

\[\begin{array}{1 1}(1)\sqrt{2}+1&(2)\sqrt{2}-1\\(3)2\sqrt{2}-2&(4)2\sqrt{2}+2\end{array}\]

Can you answer this question?

1 Answer

0 votes
Area $A= \int \limits _0^{\pi/4} (y_1-y_2)$
$\qquad= \int \limits _0^{\pi/4} (\cos x - \sin x )dx$
$\qquad= [ \sin x -(- \cos x ) ] _0^{\pi/4}$
$\qquad= (\sin \large\frac{\pi}{4} -$$ \sin 0 ) + (\cos \large\frac{\pi}{4} $$- \cos 0)$
$\qquad= (\large\frac{1}{\sqrt 2} -0)+ (\frac{1}{\sqrt 2} -1)$
$\qquad = \large\frac{2}{\sqrt 2}$$ -1$
$\qquad= \sqrt 2-1$
Hence 2 is the correct answer.
answered May 21, 2014 by meena.p

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