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?
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
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.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