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MHB Inequality involving Gaussian integral
Thanks Klaas! So if I understand correctly, I can compute the RHS using the squeeze theorem. Since $e^{x^2}$ is strictly increasing on $J = [1,2]$, with absolute minimum value $f(1) = e^{1^2} \approx 2.71$ and absolute maximum value $f(2) = e^{2^2} = e^4 \approx 54.59$. Using the comparison...- ChrisOlafsson
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- Forum: Calculus
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C
MHB Inequality involving Gaussian integral
I'm trying to solve the inequality: $$ \int \limits_0^1 e^{-x^2} \leq \int \limits_1^2 e^{x^2} dx $$I know that $\int \limits_0^1 e^{-x^2} \leq 1$, but don't see how to take it from there. Any ideas?- ChrisOlafsson
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- Gaussian Gaussian integral Inequality Integral
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- Forum: Calculus