I got this at my final calculus 1 exam. Any help with the solution?
$$ \int_{-1}^{1} \frac{x^4}{a^x+1}dx $$
Thank you!
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HINT:
Use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
and $$I+I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b[f(x)+f(a+b-x)]dx$$
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nice trick, (+1) – Alex May 19 '14 at 08:28
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1+1. An old trick that leads to$$I=\frac12\int_{-1}^1x^4\ dx.$$ – Tunk-Fey May 19 '14 at 08:49
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Thank you all for your answers! – user83081 May 19 '14 at 11:22
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Computing the even part of the integrand, one obtains $$\int_{-1}^1 \frac{x^4dx}{a^x+1}=\frac12\int_{-1}^1 \underbrace{\left[\frac{1}{a^x+1}+\frac{1}{a^{-x}+1}\right]}_{=1}x^4dx=\frac{x^5}{10}\biggl|_{-1}^1=\frac15.$$
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