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Question Number 218624 by Nicholas666 last updated on 13/Apr/25

$$$$$$Prove;{\underset{\frac{1}{e}}{\int }}^e\frac{t^2}{e^{t^2}}dt⩽1-\frac{1}{e^2}$$$$e-thebaseofnaturallogarithm$$$$$$

Answered by Nicholas666 last updated on 13/Apr/25

$${\int }_{1/e}^e\frac{t^2}{e^{t^2}}dt⩽1-\frac{1}{e^2}$$$$\Rightarrow f^{\prime }t=\frac{d}{dt}\left(t^2{e}^{-t^2}\right)=2{te}^{-t^2}+t^2(-2{te}^{-t^2})=$$$$2{te}^{-t^2}(1-t^2)$$$$\Rightarrow f^{\prime }(0.5)=2(0.5)e^{-{(0.5)}^2}=$$$$e^{-0.25}(1-0.25)=0.75e^{-25}>0$$$$\Rightarrow f^{\prime }\left(2\right)=2\left(2\right)e^{-2^2}(1-2^2)=$$$$4e^{-4}(1-4)=-12e^{-4}<0$$$$\Rightarrow f\left(1\right)=\frac{1^2}{e^{12}}=\frac{1}{e}$$$$\Rightarrow {\int }_{1/e}^e\frac{t^2}{e^{t^2}}dt⩽{\int }_{1/e}^e\frac{1}{e}dt$$$$\Rightarrow {\int }_{1/e}^e\frac{1}{e}dt=\frac{1}{e}{\int }_{1/e}^{e}1dt=\frac{1}{e}{\left[t\right]}_{1/e}^e=$$$$\frac{1}{e}(e-\frac{1}{e})=\frac{e^{2-1}}{e^2}=1-\frac{1}{e^2}$$$$\Rightarrow {\int }_{1/e}^e\frac{t^2}{e^{t^2}}dt⩽1-\frac{1}{e^2}Q.E.D$$$$$$

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