MHB Inequality involving Gaussian integral

[ChrisOlafsson]

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?

 

[I like Serena]

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]

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?

 

[I like Serena]

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)