Question

Which of the following $f\left(x\right)$ bounds the same area with corresponding $g\left(x\right)$ :

• A. $f\left(x\right)=x^2-1$ , $g\left(x\right)=1-(1-x{)}^2$ with x-axis.
• B. $f\left(x\right)=\left[x\right]{\sin }^{-1}x,$ , $g\left(x\right)=\left[2x\right]{\cos }^{-1}2x$ with x axis.
• C. $f\left(x\right)=|\sin x|$ , $g\left(x\right)={\sin }^{3}x$, where $x\in [0,10\pi ]$
• D. $f\left(x\right)=|\cos x|$ , $g\left(x\right)={\sin }^{2}x$, where $x\in [0,10\pi ]$

Answer 1

The correct option is A $f\left(x\right)=x^2-1$ , $g\left(x\right)=1-(1-x{)}^2$ with x-axis.

$f\left(x\right)=x^2-1$ and $g\left(x\right)=1-(1-x{)}^2$
$\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\frac{2}{3}-2=-\frac{4}{3}$
Area $=\frac{4}{3}$
$g\left(x\right)=1-(1-x{)}^2=2x-x^2\underset{0}{\overset{2}{\int }}g\left(x\right)dx=[x^2-\frac{x^3}{3}{]}_0^2=\frac{4}{3}$
So area will be same.


$f\left(x\right)=\left[x\right]{\sin }^{-1}x,$ , $g\left(x\right)=\left[2x\right]{\cos }^{-1}2x$ with x axis.
For,
$f\left(x\right)=\left[x\right]{\sin }^{-1}x,x\in [-1,1]$
$I_1=\underset{-1}{\overset{1}{\int }}f\left(x\right)dx=\underset{-1}{\overset{0}{\int }}-{\sin }^{-1}xdx\Rightarrow I_1=\underset{0}{\overset{-1}{\int }}{\sin }^{-1}xdx$
Putting ${\sin }^{-1}x=y\Rightarrow dx=\cos ydy$
$\Rightarrow I_1=\underset{0}{\overset{-\frac{\pi }{2}}{\int }}y\cos ydy\Rightarrow I_1=[y\sin y+\cos y{]}_0^{-\frac{\pi }{2}}\Rightarrow I_1=(\frac{\pi }{2}-1)$

$g\left(x\right)=\left[2x\right]{\cos }^{-1}2x,x\in [\frac{-1}{2},\frac{1}{2}]$
$I_2=\underset{-\frac{1}{2}}{\overset{\frac{1}{2}}{\int }}g\left(x\right)dx=-\underset{-1/2}{\overset{0}{\int }}{\cos }^{-1}2xdx$
Putting ${\cos }^{-1}2x=y\Rightarrow dx=-\frac{1}{2}\sin ydy$
$\Rightarrow I_2=\frac{1}{2}\underset{\pi }{\overset{\frac{\pi }{2}}{\int }}y\sin ydy\Rightarrow 2I_2=[-y\cos y+\sin y{]}_{\pi }^{\frac{\pi }{2}}\Rightarrow 2I_2=1-\pi \Rightarrow I_2=\frac{1-\pi }{2}$
Area under $g\left(x\right)$ will be $\frac{\pi -1}{2}$
Both areas are different.

$f\left(x\right)=|\sin x|$ , $g\left(x\right)={\sin }^{3}x$, where $x\in [0,10\pi ]$
From the graph we can see that there areas cannot be equal.


$f\left(x\right)=|\cos x|$ , $g\left(x\right)={\sin }^{2}x$, where $x\in [0,10\pi ]$
$\underset{0}{\overset{10\pi }{\int }}f\left(x\right)dx=20\times \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos xdx=20$
$\underset{0}{\overset{10\pi }{\int }}g\left(x\right)dx=20\times \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2}xdx\Rightarrow \underset{0}{\overset{10\pi }{\int }}g\left(x\right)dx=5\pi$
So there area will not be equal