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Just like in the title, one of the problems given in our homework is to prove that for every real $x$ the following equality is true: $$ \int_{-x}^{x} \frac{t^2 e^t}{e^t + 1} \, dt = \frac{x^3}{3}$$

I've been facing this for 2 days straight now, and am absolutely dumbstruck on how to do this. We aren't allowed to calculate the antiderivative of this problem.

The only hint we were given is to think of an operation that will cancel out the integral on the left side of the equality. But even after rereading out learning material again and again, I simply couldn't find anything helpful.

However, I would much rather getting a hint rather than the solution if possible.

Thanks in advance!

EDIT: Even after reading the solution written below, I still feel like something's not clicking in my head. If possible, a detailed explanation would be ideal (and thanks in advance for going through the trouble for me).

Emanuel L
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1 Answers1

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Hint 1: Fundamental theorem of calculus

Hint 2: Let $F(x)=\displaystyle\int^x_{-x}\dfrac{t^2e^t}{e^t+1}~dt$. What is $F(0)$?.

Edit. Let's me provide the exact solution below in case you need.

Hint 1: We apply fundamental theorem of calculus on $F$. That is, $$F'(x)=\dfrac{x^2e^x}{e^x+1}\dfrac{dx}{dx}-\dfrac{(-x)^2e^{-x}}{e^{-x}+1}\dfrac{d(-x)}{dx}=x^2\left(\dfrac{e^x}{e+1}+\dfrac{e^{-x}}{e^{-x}+1}\times\dfrac{e^x}{e^x}\right)=x^2$$Hence, $F(x)=\dfrac{x^3}{3}+\text{constant}$. By Hint 2, $F(0)=0$, so you have shown that $F(x)=\dfrac{x^3}{3}$.

Angae MT
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  • It occurs to me that I'm dumb and still don't have any clue on what to do... $F(0)=0$ because it is a single point, but what do I infer from this? – Emanuel L Mar 29 '24 at 22:01
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    Hint 1 provides you the antiderivative of $F$, so $F(x)=\dfrac{x^3}{3}+C$ but how do you know which $C$ is? You need some specific value of $F(x)$. – Angae MT Mar 29 '24 at 22:03
  • @EmanuelL Compute $F'(x)$ via hint 1. – Gary Mar 29 '24 at 22:04
  • I understand that what I am searching for is: $F(x)-F(-x)$. How were you able to conclude that $(x^3)/3 + C$ is the antiderivative? – Emanuel L Mar 29 '24 at 22:21
  • You are not searching for $F(x)-F(-x)$. $F(x)$ is exactly the integral you want. Your goal is to show $F(x)=\dfrac{x^3}{3}$ under my definition. – Angae MT Mar 29 '24 at 22:22
  • It is wrong to provide a complete solution to a homework problem. OP even said a complete solution is not sought after. – geetha290krm Mar 29 '24 at 23:12
  • Even after reading the solution, I honestly don't get it... I feel stupid or that I am missing fundamental knowledge or understanding... – Emanuel L Mar 29 '24 at 23:32
  • @geetba290krm That's why I give a spoiler alert over the solution, you can choose to not see this, but in case you need, there is extra guidance to them. I don't think giving soln to OP is harming him. – Angae MT Mar 29 '24 at 23:54
  • Hopefully I'm not mistaken, but I think its the definition of $F(x)$ that confuses me, and how you got it's derivative. – Emanuel L Mar 29 '24 at 23:56
  • I have a last silly question. You say that if we take the derivative of $F(x)=G(x)-G(-x)$ we get $F'(x)=G'(x)-G'(-x)(-1)$. Does the $-1$ come from the chain rule? – Emanuel L Mar 30 '24 at 00:11
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    @EmanuelL Yes, it comes from the chain rule. You may also take a look at the Leibniz integral rule. – Gary Mar 30 '24 at 22:36