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- #1

#### ChrisOlafsson

##### New member

- Mar 8, 2020

- 2

$$

\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?

- Thread starter ChrisOlafsson
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- Thread starter
- #1

- Mar 8, 2020

- 2

$$

\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?

- Admin
- #2

- Mar 5, 2012

- 8,736

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...

- Thread starter
- #3

- Mar 8, 2020

- 2

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?

- Admin
- #4

- Mar 5, 2012

- 8,736

Yep.

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?