Question

If $I_1={\int }_0^1{e}^{-x}{\cos }^{2}xdx,I_2={\int }_0^1{e}^{-x^2}{\cos }^{2}xdx$
$I_3={\int }_0^1{e}^{-x^2}dx,I_4={\int }_0^1{e}^{-x^2}dx$ then

• A. $I_1>I_2>I_3>I_4$
• B. $I_4>I_3>I_2>I_1$
• C. $I_4>I_1>I_3>I_2$
• D. $I_4>I_3>I_1>I_2$

Answer 1

The correct option is B $I_4>I_3>I_2>I_1$

When $0<x<1$,

$x^2<x$

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

$\Rightarrow e^{-x^2}>e^{-x}$
Multiplying both sides by ${\cos }^{2}x$ we get

$e^{-x^2}{\cos }^{2}x>e^{-x}{\cos }^{2}x$

$\Rightarrow {\int }_0^1{e}^{-x^2}{\cos }^{2}xdx>{\int }_0^1{e}^{-x}{\cos }^{2}xdx$

$\Rightarrow I_2>I_1$
Similarly,

since,$0<{\cos }^{2}x<1$

since $0<\frac{co{s}^{2}x}{e^{x^2}}<e^{-x^2}\Rightarrow I_3>I_2\Rightarrow I_3>I_2>I_1$
And
$e^{{\frac{x}{2}}^2}<e^{x^2}\Rightarrow \frac{1}{e^{{\frac{x}{2}}^2}}>\frac{1}{e^{x^2}}\Rightarrow I_4>I_3\Rightarrow I_4>I_3>I_2>I_1$