Solution from greg1313:Solution from lfdahl:

[anemone]

I would like to say a big thank you to MarkFL, who stood in for me during the last month to take care of the POTW duty while I was away in another country.Here is this week's POTW:

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Let $b\gt a \gt 0$.

Prove that

$$\int_{a}^{b} (x^2+1)e^{-x^2} \,dx\ge e^{-a^2}-e^{-b^2}$$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. MarkFL
2. Ackbach
3. greg1313
4. lfdahl

Solution from MarkFL:

Spoiler

Since, for all real $x$ we have:

$$x^2+1\ge 2x$$

We may state:

$$\int_a^b (x^2+1)e^{-x^2}\,dx\ge\int_a^b 2xe^{-x^2}\,dx=e^{-a^2}-e^{-b^2}$$

Alternate solution from Ackbach:

Spoiler

Note that
\begin{align*}
\int_a^b(x^2+1)e^{-x^2} \, dx&=\int_a^b(x^2-2x+1)e^{-x^2} \, dx-\int_a^b(-2x)e^{-x^2} \, dx \\
&=\int_a^b(x-1)^2e^{-x^2} \, dx-\left[e^{-b^2}-e^{-a^2}\right].
\end{align*}
Since $(x-1)^2e^{-x^2}\ge 0$ and $0<a<b$, it follows that $\displaystyle \int_a^b(x-1)^2e^{-x^2} \, dx> 0$, and therefore
$$\int_a^b(x^2+1)e^{-x^2} \, dx>-\left[e^{-b^2}-e^{-a^2}\right]=e^{-a^2}-e^{-b^2}.$$
QED.