# Proof area of {$(x,y) R^2 | 0 \le y < e^x$ and $0\le x< h$} is $e^h-1$

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.

• look up the fundamental theorem of calculus – wilkersmon Dec 6 '18 at 11:08
• It is a geomtrical serie, the sum is equal to $$h/n\sum\limits_{i = 0}^{n - 1} {{e^{ih/n}}} = \left( {h/n} \right)\frac{{1 - {e^{(h + \frac{h}{n})}}}}{{1 - {e^{h/n}}}}$$ – Gustave Dec 6 '18 at 11:35

$$\sum_{i=0}^n e^\frac{hi}{n}$$ 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
$$\frac hn \sum_{i=0}^n e^\frac{hi}{n} \rightarrow e^h-1$$