Question

If $I_1={\int }_0^1{e}^{-x}{\cos }^{2}xdx$; $I_2={\int }_0^1{e}^{-x^2}{\cos }^{2}xdx$ and $I_3={\int }_0^1{e}^{-x^3}dx$; then

• A. $I_2>I_3>I_1$
• B. $I_3>I_1>I_2$
• C. $I_2>I_1>I_3$
• D. $I_3>I_2>I_1$

Answer 1

The correct option is D $I_3>I_2>I_1$

Given:

$I_1={\int }_0^1{e}^{-x}{\cos }^{2}xdx$;

$I_2={\int }_0^1{e}^{-x^2}{\cos }^{2}xdx$ and

$I_3={\int }_0^1{e}^{-x^3}dx$

For $x\in (0,1),x>x^2\Rightarrow -x<-x^2$

$\Rightarrow x^2>x^3$

$\Rightarrow -x^2<-x^3$

$\Rightarrow e^{-x^2}<e^{-x^3}$

and $e^{-x}<e^{-x^2}$

$\Rightarrow e^{-x}<e^{-x^2}<e^{-x^3}$

$\Rightarrow e^{-x^3}>e^{-x^2}>e^{-x}$

$\Rightarrow I_3>I_2>I_1$

Green line denotes $f\left(x\right)=e^{-x}{\cos }^{2}x$

Blue line denotes $g\left(x\right)=e^{-x^2}{\cos }^{2}x$

Red line denotes $h\left(x\right)=e^{-x^3}$

Also, from the graph we get the same result.

Hence, option D is correct.