Showing $ \displaystyle\int_{t-1}^t \log{x}\mathrm dx < \log{t} $

Question

How can I show that

$$ \int_{t-1}^t \log{x}dx < \log{t} $$

I can calculate the integral of $\log{x}$ as, $x\log{x}-x$. But after calculating, definite integral and re-aranging, I am not able to get the desired result. Is it true that the above inequality holds true for any monotonic function ?

Answer 1

For each $x\in[t-1,t)$ , $\log(x)<\log(t)$ and $\log$ is a continuous function. Therefore\begin{align}\int_{t-1}^t\log(x)\,\mathrm dx&<\int_{t-1}^t\log(t)\,\mathrm dx\\&=\bigl(t-(t-1)\bigr)\log(t)\\&=\log(t).\end{align}