Prove that Riemann Sum is larger Let $ f:[a,b] \rightarrow \mathbb{R}$ be a differentiable function. Show that if $ P = (x_0, x_1, ..., x_n) $ is a partition of [a,b] then
$$ U(P, f') := \sum_{j=1}^n M_j \Delta x_j \ge f(b) - f(a) $$ where $$ M_j := sup\{f'(t) : t \in [x_{j-1}, x_j] \}$$
How do I go about proving this? I can draw a graph of it and clearly the Riemann sum is always larger than the definite integral but I'm not sure how to go about proving it otherwise.

If $f$ is continuously differentiable, then by fundamental theorem of calculus we have $\int_{a}^{b}f' = f(b) - f(a)$; the result follows from the fact that Darboux integration is equivalent to Riemann integration and that $\inf U(f'; P) = \int_{a}^{b}f'$ by definition. 
If $f$ is simply differentiable, then by mean-value theorem we have
$f(x_{i}) - f(x_{i-1}) \leq (x_{i}-x_{i-1})M_{i}$ for all $i$; summing over all $i$ gives the desired result.
