Question

Consider the integrals $I_1={\int }_0^1{e}^{-x}co{s}^{2}xdx$
$I_2={\int }_0^1{e}^{-x^2}co{s}^{2}xdx,I_3={\int }_0^1{e}^{-x^2}dx$ and
$I_4={\int }_0^1{e}^{-\left(1/2\right)x^2}dx$. The greatest of these integrals $I$ is

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

Answer 1

Answer: (D)

$I_4$

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