Prove a set has an area of $\;e^h - 1$

Question

Prove through appropriate estimations with simple sets that the set

$$E =\{(x, y) ∈ ℝ^2 ∣∣ 0 ≤ y < e^x \quad \& \quad 0 ≤ x < h\}$$ has an area of $$e^h - 1.$$

I believe I have to represent the sets as Riemann sums of the upper and lower limits, which I have done the following way:

Lower: $$|F| =\sum_{k=1}^n e^\frac{h(k-1)}{n} \frac{h}{n}$$

Upper: $$|G| =\sum_{k=1}^n e^\frac{hk}{n} \frac{h}{n}$$

Now my question is: how do I simplify $$\sum_{k=1}^n e^\frac{h(k-1)}{n} \frac{h}{n}$$ and $$\sum_{k=1}^n e^\frac{hk}{n} \frac{h}{n}$$ to expressions where I can show that $$|F| → e^h - 1$$ $$|G| → e^h - 1$$ when n → infinity?$$$$ When I type in my two Riemann sums into Wolfram Alpha I don't get any good simplifications of the sums. What am I doing wrong? How would you approach this problem?

Answer 1

Substantial hint:

In $$\sum_{k=1}^n e^\frac{hk}{n} \frac{h}{n},$$ I'd notice that $h/n$ is a constant that could be taken out of the sum to get $$ \frac{h}{n} \sum_{k=1}^n e^\frac{hk}{n} $$

Then I'd notice that $$ e^\frac{hk}{n} = \left(e^\frac{h}{n}\right)^k $$ and perhaps rename the constant $e^\frac{h}{n}$ with a single letter like $b$, and see whether the resulting expression brought anything to mind.

In short: I'd spend some time doing some algebra, and avoid rushing to Wolfram Alpha every time I encountered an expression with more than a few symbols.

Answer 2

Since you tagged "Calculus" your question, let us do it:

$$\lim_{n\to\infty}\;h\frac1n\sum_{k=1}^ne^{h\frac kn}=h\int_0^1e^{hx}dx=\ldots$$

Answer 3

Found a way to solve it. Realized I introduce a new variable $c = e^\frac{h}{n}$ and write the sum as a geometrical sum instead. Made things clearer.