Here is the question I am trying to answer:
Let $f:[0,1] \to\Bbb R$ be a Riemann integrable function with $f \ge c>0$. Prove that $$\int_0^1\ln(f(x))\ dx\le \ln\left(\int_0^1 f(x)\ dx\right).$$
I understand how to prove two integrals are equal by showing that their upper and lower Darboux sums are equal and that they converge to the same definite integral. But, I don't understand how to prove the less than or equal to part.
Ideas I've thought about include: integration by parts, improper integrals, partitions.
Does anyone know how to prove this?
Jensen's inequality holds for an interval $[a,b]$, an integrable nonnegative function $f$ from $[a,b]$ to the real line and a convex function $\varphi$, and it states: $$\varphi\left(\int_a^bf(x) dx\right)\leq \int_a^b\varphi\left(f(x)\right)dx.$$ This does not apply to the $\log(x)$ function because it is concave away from zero, but this means the function $(-\log(x))$ is convex away from zero. Applying Jensen's: $$-\log\left(\int_a^bf(x) dx\right)\leq -\int_a^b \log\left(f(x)\right)dx,$$ which yields $$\int_a^b \log\left(f(x)\right)dx \leq \log\left(\int_a^bf(x) dx\right).$$ Zero is not a problem point because of your assumptions on $f$.