Question

The area bounded by $y=x^2+2$ and $y=2|x|-\cos \pi x$ is

• A. $\frac{1}{3}\text{sq. units}$
• B. $\frac{2}{3}\text{sq. units}$
• C. $\frac{4}{3}\text{sq. units}$
• D. $\frac{8}{3}\text{sq. units}$

Answer 1

The correct option is D $\frac{8}{3}\text{sq. units}$


$y=x^2+2$, $y=2|x|-\cos \pi x$
Finding the intersection point:
$x^2+2=2|x|-\cos \pi x$
$\Rightarrow |x{|}^2+1+1-2|x|=-\cos \pi x$
$\Rightarrow (|x|-1{)}^2+1=-\cos \pi x$
Range of $(|x|-1{)}^2+1$ is $[1,\infty )$
Range of $-\cos \pi x$ is $[-1,1]$
The equality is true only when $(|x|-1{)}^2+1=-\cos \pi x=1$
$\Rightarrow x=\pm 1$

Area under curve $=\underset{-1}{\overset{1}{\int }}\left[\right(x^2+2)-(2|x|-\cos \pi x\left)\right]dx$
$=2\underset{0}{\overset{1}{\int }}\left[\right(x^2+2)-2x+\cos \pi x]dx$
$=2{[\frac{x^3}{3}+2x-x^2+\frac{1}{\pi }\sin \pi x]}_0^1$
$=2[\frac{1}{3}+2-1+0]$
$=\frac{8}{3}\text{sq. units}$