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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.

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

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

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