I'm supposed to prove that the area of {$(x,y) R^2 | 0 \le y < e^x$ and $0\le x< h$} is $e^h-1$
I was going to try to make it a function and calculate it using a Riemanns sum.
That led me to
$F(x) = e^2 = y$
Assuming n rectangles with the width $h/n$ and height $e^\frac{hi}{n}$ That got me to the sum and now I'm stuck at
$$\frac hn \sum_{i=0}^n e^\frac{hi}{n} $$
How should I proceed? The proof is supposed to use simple sets, e.i. not supposed to use an integral.
$\sum_{i=0}^n e^\frac{hi}{n}$ is a geometric series with ratio $e^{\frac hn}$ so
$\sum_{i=0}^n e^\frac{hi}{n} = \frac{e^{\frac{h(n+1)}{n}}-1}{e^{\frac hn}-1}$
As $n \rightarrow \infty$, we have $e^{\frac{h(n+1)}{n}}-1 \rightarrow e^h-1$ and $e^{\frac hn}-1 \rightarrow \frac hn$ so
$\frac hn \sum_{i=0}^n e^\frac{hi}{n} \rightarrow e^h-1$