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Inequality involving Gaussian integral

ChrisOlafsson

New member
Mar 8, 2020
2
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
Hi and welcome to MHB!

Note that:
$$(b-a)\min_{a\le x\le b}f(x)\quad\le\quad\int_a^b f(x)\,dx \quad\le\quad (b-a)\max_{a\le x\le b}f(x)$$

If we apply that to the left side of your inequality, it gives us what you already know.
That leaves the right side...
 

ChrisOlafsson

New member
Mar 8, 2020
2
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 theorem,
$
e^{1} \leq \int \limits_1^2 e^{x^2} \leq e^{2^2}
$
and since since this area is greater than 1, consequently, the inequality holds.

Is this what you meant?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
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 theorem,
$
e^{1} \leq \int \limits_1^2 e^{x^2} \leq e^{2^2}
$
and since since this area is greater than 1, consequently, the inequality holds.

Is this what you meant?
Yep. (Nod)