Calculate: $\int_{-2}^{2}\frac{x^2}{e^x+1}~dx$

Question

$$\int_{-2}^{2}\frac{x^2}{e^x+1}~dx$$ Options:

a) $\; 0$

b) $\; \frac{e-2}{e}$

c) $\; \frac{8}{3}$

d) $\; \frac{e-1}{e}$

e) $\; \frac{16}{3}$

Note: It's high school level

What I tried:

Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. As this question appears to be homework, please consider reading this page for information about asking effective homework-related questions. Cheers! — David Raveh, Jul 20 '23 at 21:27
This graph indicates that the value of the integral should be kinda close to 2.5 and 8/3 is the only choice that fills that criterion. Note: While I generated the graph using Desmos for illustrative purposes, don't think it's too hard to come up with the picture without software. — Mark McClure, Jul 20 '23 at 21:43
If the problem is multiple choice and targeted at high schoolers, then you should absolutely consider a geometric approach. In this problem, it literally takes a minute or two. That's probably the point from the problem writer's perspective. — Mark McClure, Jul 20 '23 at 21:59

score 6 · Accepted Answer · answered Jul 20 '23 at 23:33

6

Let $x \mapsto -x$ (this is the same as letting $u = -x $ then renaming $u$ as $x$).

$\displaystyle I = \int_{-2}^{2}\frac{x^2}{e^x+1}~dx = -\int_{2}^{-2}\frac{x^2}{e^{-x}+1}~dx = \int_{-2}^{2}\frac{x^2}{e^{-x}+1}~dx$

Adding the first and last integrals together and using $\frac{1}{y+1}+\frac{1}{y^{-1}+1} = 1$:

$\displaystyle 2I = \int_{-2}^{2}\frac{x^2}{e^x+1}~dx + \int_{-2}^{2}\frac{x^2}{e^{-x}+1}~dx = \int_{-2}^{2} x^2 dx = \frac{16}{3}$

So that $\displaystyle {I = \frac{8}{3}}$.

answered Jul 20 '23 at 23:33

NoName

I recently learned that this trick is called the King's property – David Raveh Jul 21 '23 at 03:16
@DavidRaveh It has discrete versions:
$$\displaystyle \sum_{k=a}^{b}f(k) = \sum_{k=a}^{b}f(a+b-k)$$

$$\displaystyle \prod_{k=a}^{b}f(k) = \prod_{k=a}^{b}f(a+b-k)$$
– NoName Jul 21 '23 at 03:32
This solution was already in the link to AoPS provided above by @TymaGaidash, but it is good to have it here. – Anne Bauval Jul 21 '23 at 04:16
1

@AnneBauval It's a standard trick – NoName Jul 21 '23 at 04:31
I am used to it. – Anne Bauval Jul 21 '23 at 04:36

1 Answers1