Equation with integral: int[-a,a] [(x^4)/(e^x+1)] dx = -32/5

[kougami]

I have this integral:

\(\displaystyle \displaystyle \int _{-a}^a\frac{x^4}{e^x+1}\:dx=-\frac{32}{5}\)

I need to find a (∈R), but I have no idea how to begin. Can you help me? Thank you!

 

[tkhunny]

I have this integral:

\(\displaystyle \displaystyle \int _{-a}^a\frac{x^4}{e^x+1}\:dx=-\frac{32}{5}\)

I need to find a (∈R), but I have no idea how to begin. Can you help me? Thank you!

Is \(\displaystyle \dfrac{x^4}{e^x+1}\) EVER negative?

 

[kougami]

Is \(\displaystyle \dfrac{x^4}{e^x+1}\) EVER negative?

no, but if a it's negative, integral will be:
\(\displaystyle \displaystyle \int _a^{-a}\frac{x^4}{e^x+1}\:dx=-\int _{-a}^a\frac{x^4}{e^x+1}\:dx\)
So, I think that a<0

 

[tkhunny]

no, but if a it's negative, integral will be:
\(\displaystyle \displaystyle \int _a^{-a}\frac{x^4}{e^x+1}\:dx=-\int _{-a}^a\frac{x^4}{e^x+1}\:dx\)
So, I think that a<0

True. I missed that, earlier. I was just coming in to check when I thought of it. Good work.

 

[tkhunny]

no, but if a it's negative, integral will be:
\(\displaystyle \displaystyle \int _a^{-a}\frac{x^4}{e^x+1}\:dx=-\int _{-a}^a\frac{x^4}{e^x+1}\:dx\)
So, I think that a<0

ONE application of Integration by Parts seems to be helpful. It may lead somewhere.

\(\displaystyle \int\dfrac{1}{e^{x} + 1} = x - \ln(e^{x} + 1) + C\)

This leads to three things:

1) \(\displaystyle [x - \ln(e^{x} + 1)]\cdot x^{4}\) - Which may surprise you as evaluating to a constant on ANY interval symmetric about the Origin.

2) An odd function - that you can just ignore on any interval symmetric about the Origin.

3) Another integral that may suggest to you where to look next.

Are we getting anywhere?

 

[Steven G]

no, but if a it's negative, integral will be:
\(\displaystyle \displaystyle \int _a^{-a}\frac{x^4}{e^x+1}\:dx=-\int _{-a}^a\frac{x^4}{e^x+1}\:dx\)
So, I think that a<0

you can do that with any integral, whether a is positive or negative.