Question

If I is the greatest of the definite integrals
$I_1={\int }_0^1{e}^{-x}co{s}^{2}xdx,I_2={\int }_0^1{e}^{-x^2}co{s}^{2}xdx$
$I_3={\int }_0^1{e}^{-x^2}dx,I_4={\int }_0^1{e}^{\frac{-x^2}{2}}dx,then$

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

Answer 1

The correct option is D

$I=I_4$

For 0 < x < 1, we have $\frac{1}{2}{x}^2<x^2<x$
$\Rightarrow -x^2>-x,sothat{e}^{-x^2}<e^{-x},$
$Hence{\int }_0^1{e}^{-x^2}co{s}^{2}xdx>{\int }_0^1{e}^{-x}co{s}^{2}xdx$
$Alsoco{s}^{2}x\le 1$
$Therefore{\int }_0^1{e}^{-x^2}co{s}^{2}xdx\le {\int }_0^1{e}^{-x^2}dx<{\int }_0^1{e}^{\frac{-x^2}{2}}dx=I_4$
Hence $I_4$ is the greatest integral