Question

Solve this:
$$\begin{gathered} If{I}_1={\int }_0^1{e}^{-x}{\cos }^{2}xdx, \\ {I}_2={\int }_0^1{e^{-x}}^2{\cos }^{2}xdxand \\ {I}_3={\int }_0^1{e}^{-x^3}dx;then: \\ 1.I_3>I_1>I_2 \\ 2.I_2>I_3>I_1 \\ 3.I_3>I_2>I_1 \\ 4.I_2>I_1>I_3 \end{gathered}$$

Answer 1

Dear student
$$\begin{gathered} {\mathrm{I}}_1={\int }_0^1{\mathrm{e}}^{-\mathrm{x}}{\cos }^2\mathrm{x}d\mathrm{x} \\ {\mathrm{I}}_2={\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^2}{\cos }^2\mathrm{x}d\mathrm{x} \\ {\mathrm{I}}_3={\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^3}d\mathrm{x} \\ \mathrm{For}0<\mathrm{x}<1,{\mathrm{x}}^2<{\mathrm{x}}^2<\mathrm{x} \\ \mathrm{Consider}, \\ {\mathrm{x}}^2<\mathrm{x} \\ -{\mathrm{x}}^2>-\mathrm{x} \\ \Rightarrow {\mathrm{e}}^{-{\mathrm{x}}^2}>{\mathrm{e}}^{-\mathrm{x}} \\ \Rightarrow {\mathrm{e}}^{-{\mathrm{x}}^2}{\cos }^2\mathrm{x}>{\mathrm{e}}^{-\mathrm{x}}{\cos }^2\mathrm{x} \\ \Rightarrow {\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^2}{\cos }^2\mathrm{x}\mathrm{dx}>{\int }_0^1{\mathrm{e}}^{-\mathrm{x}}{\cos }^2\mathrm{x}\mathrm{dx} \\ \mathrm{and}{\cos }^2\mathrm{x}\le 1 \\ \Rightarrow {\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^2}{\cos }^2\mathrm{x}\mathrm{dx}<{\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^3}d\mathrm{x} \\ \mathrm{So},{\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^3}d\mathrm{x}>{\int }_0^1{\mathrm{e}}^{-{\mathrm{x}}^2}{\cos }^2\mathrm{x}\mathrm{dx}>{\int }_0^1{\mathrm{e}}^{-\mathrm{x}}{\cos }^2\mathrm{x}\mathrm{dx} \\ {\mathrm{I}}_3>{\mathrm{I}}_2>{\mathrm{I}}_1 \end{gathered}$$
Regards