If I is the greatest of the definite integrals  

$\begin{array}{l}{I}_1={\int }_0^1 e^{-x}{\cos }^{2}x\mathrm{d}x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{I}_2={\int }_0^1 e^{-x^2}{\cos }^{2}x\mathrm{d}x,\\ I_3={\int }_0^1 e^{-x^2}\mathrm{d}x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{I}_4={\int }_0^1 e^{-x^2/2}\mathrm{d}x\end{array}$

then

For $0<x<1$, we have $(1/2)x^2<x^2<x$ i.e., $-x^2>-x$ so that $e^{-x^2}>e^{-x}$

Hence ${\int }_0^1 e^{-x^2}{\cos }^{2}x\mathrm{d}x>$

${\int }_0^1 e^{-x}{\cos }^{2}xdx$. Also  ${\cos }^{2}x\le 1$, therefore

${\int }_0^1 e^{-x^2}{\cos }^{2}x\mathrm{d}x\le {\int }_0^1 e^{-x^2}\mathrm{d}x<{\int }_0^1 e^{-x^2/2}\mathrm{d}x=I_4.$

Hence $I_4$ is the greatest integral.