Question

The area bounded by $y=\cos x,x=-\frac{\pi }{2},x=2\pi$ and the axis of $x$ is square units is

• A. 5
• B. 2
• C. 4
• D. 1

Answer 1

The correct option is A 5

Split $(-\frac{\pi }{2},2\pi )$ to $(-\frac{\pi }{2},\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{2}).(\frac{3\pi }{2},2\pi )$

$y$ is $+$ ve in the $1st$ and $3rd$ intervals and it is $-$ve in the $2nd$ interval.

$A={\int }_{-\pi /2}^{\pi /2}ydx+\left|{\int }_{\pi /2}^{3\pi /2}ydx\right|+{\int }_{3\pi /2}^{2\pi }ydx$

$A={\left[\sin x\right]}_{-\pi /2}^{\pi /2}+\left|{\left[\sin x\right]}_{\pi /2}^{3\pi /2}\right|\Rightarrow A=(1+1)+[0-\left(1\right)].$

$A=2+2+1=5$ sq. units