# Comparing two definite integrals

I need help proving this inequality.

$$\int_{1}^{2} \left(\frac{x+1}{2} \right)^{x+1} \ dx < \int_{1}^{2} x^x \ dx$$

Normally I would define $f(x) = \left(\displaystyle\frac{x+1}{2} \right)^{x+1}$ and $g(x) = x^x$ and try to compare those two.
Any suggestions?

-
Evaluate each integrals separately. For f(x) and g(x), evaluate using numerical methods. –  Neigyl Noval Jan 23 '11 at 11:49

HINT: In $[1,2]$ we have that $x^x > (\frac{x+1}{2})^{x+1}$. Try to show this by finding the minimum of $h(x) = x\log(x)-(x+1)\log(\frac{x+1}{2})$.