Recall that we wish to compute the area of

\begin{gather*} \big\{\ (x,y)\ \big|\ 0\le y\le e^x,\ 0\le x\le 1\ \big\} \end{gather*}

and that our strategy is to approximate this area by the area of a union of a large number of very thin rectangles, and then take the limit as the number of rectangles goes to infinity. In Example 1.1.1, we used just four rectangles. Now we'll consider a general number of rectangles, that we'll call \(n\text{.}\) Then we'll take the limit \(n\rightarrow\infty\text{.}\) So

The area we want is exactly the sum of the areas of all of the thin strips.

Each of these strips is almost, but not quite, a rectangle. As in Example 1.1.1, the only problem is that the top is not horizontal. So we approximate each strip by a rectangle, just by levelling off the top. Again, we have to make a choice — at what height do we level off the top?

Consider, for example, the leftmost strip. On this strip, \(x\) runs from \(0\) to \(\frac{1}{n}\text{.}\) As \(x\) runs from \(0\) to \(\frac{1}{n}\text{,}\) the height \(y\) runs from \(e^0\) to \(e^{\frac{1}{n}}\text{.}\) It would be reasonable to choose the height of the approximating rectangle to be somewhere between \(e^0\) and \(e^{\frac{1}{n}}\text{.}\) Which height should we choose?

Well, as we said in Example 1.1.1, it doesn't matter. We shall shortly take the limit \(n\rightarrow\infty\) and, in that limit, all of those different choices give exactly the same final answer. We won't justify that statement in this example, but there will be an (optional) section shortly that provides the justification. For this example we just, arbitrarily, choose the height of each rectangle to be the height of the graph \(y=e^x\) at the smallest value of \(x\) in the corresponding strip 4. The figure on the left below shows the approximating rectangles when \(n=4\) and the figure on the right shows the approximating rectangles when \(n=8\text{.}\)

Now we compute the approximating area when there are \(n\) strips.

The total area of all of the approximating rectangles is

\begin{align*} \text{Total approximating area} &= \frac{1}{n}e^0 + \frac{1}{n}e^{\frac{1}{n}} + \frac{1}{n}e^{\frac{2}{n}} + \frac{1}{n}e^{\frac{3}{n}} + \cdots + \frac{1}{n}e^{\frac{(n-1)}{n}}\\ &= \frac{1}{n}\Big( 1+ e^{\frac{1}{n}} +e^{\frac{2}{n}}+e^{\frac{3}{n}} +\cdots+ e^{\frac{(n-1)}{n}}\Big) \end{align*}

Now the sum in the brackets might look a little intimidating because of all the exponentials, but it actually has a pretty simple structure that can be easily seen if we rename \(e^{\frac{1}{n}}=r\text{.}\) Then

So

\begin{align*} \text{Total approximating area} &= \frac{1}{n}\left( 1+ r +r^2 +\cdots+ r^{n-1}\right) \end{align*}

The sum in brackets is known as a geometric sum and satisfies a nice simple formula:

\begin{gather*} 1+ r +r^2 +\cdots+ r^{n-1} =\frac{r^n-1}{r-1} \qquad\text{provided $r\ne 1$} \end{gather*}

The derivation of the above formula is not too difficult. So let's derive it in a little aside.

Geometric sum

Denote the sum as

\begin{align*} S& =1 + r + r^2 + \cdots + r^{n-1} \end{align*}

Notice that if we multiply the whole sum by \(r\) we get back almost the same thing:

\begin{align*} &rS\\ &= r\left(1+ r +\cdots+ r^{n-1}\right)\\ &= r+ r^2+ r^3 +\cdots+ r^n \end{align*}

This right hand side differs from the original sum \(S\) only in that

That is

Moving this around a little gives

as required. Notice that the last step in the manipulations only works providing \(r \neq 1\) (otherwise we are dividing by zero).

Now we can go back to our area approximation armed with the above result about geometric sums.

\begin{align*} \text{Total approximating area} &= \frac{1}{n}\left( 1+ r +r^2 +\cdots+ r^{n-1}\right)\\ &= \frac{1}{n} \frac{r^n-1}{r-1} \qquad\qquad \text{remember that $r=e^{1/n}$}\\ &= \frac{1}{n} \frac{e^{n/n} - 1}{e^{1/n}-1}\\ &= \frac{1}{n} \frac{e - 1}{e^{1/n}-1} \end{align*}

To get the exact area 5 all we need to do is make the approximation better and better by taking the limit \(n\rightarrow \infty\text{.}\) The limit will look more familiar if we rename \(\frac{1}{n}\) to \(X\text{.}\) As \(n\) tends to infinity, \(X\) tends to \(0\text{,}\) so

\begin{align*} \text{Area}&=\lim_{n\rightarrow\infty} \frac{1}{n}\ \frac{e-1}{e^{1/n}-1}\\ &=(e-1)\lim_{n\rightarrow\infty} \frac{1/n}{e^{1/n}-1}\\ &=(e-1)\lim_{X\rightarrow 0} \frac{X}{e^X-1} &\text{(with $X=\frac{1}{n}$)} \end{align*}

Examining this limit we see that both numerator and denominator tend to zero as \(X\to 0\text{,}\) and so we cannot evaluate this limit by computing the limits of the numerator and denominator separately and then dividing the results. Despite this, the limit is not too hard to evaluate; here we give two ways:

\begin{align*} \lim_{X\rightarrow 0} \frac{X}{e^X-1} &=\lim_{X\rightarrow 0} \frac{\dfrac{d}{dX}X}{\dfrac{d}{dX}(e^X-1)} =\lim_{X\rightarrow 0} \frac{1}{e^X}=1 \end{align*}

\begin{align*} \lim_{X\rightarrow 0} \frac{X}{e^X-1} &= \left[\lim_{X\rightarrow 0} \frac{e^X-1}{X} \right]^{-1} \end{align*}

\begin{align*} \lim_{X\rightarrow 0} \frac{e^X-1}{X} &= \lim_{X\rightarrow 0} \frac{e^X-e^0}{X-0} \end{align*}

\begin{align*} \lim_{X\rightarrow 0} \frac{X}{e^X-1} &=\left[\lim_{X\rightarrow 0}\, \frac{e^X-e^0}{X-0} \right]^{-1}\\ &=\left[\dfrac{d}{dX}e^X\Big|_{X=0} \right]^{-1}\\ &=\left[e^X\big|_{X=0}\right]^{-1}\\ &=1 \end{align*}

So, after this short aside into limits, we may now conclude that

\begin{align*} \text{Area} &=(e-1)\lim_{X\rightarrow 0} \frac{X}{e^X-1}\\ &=e-1 \end{align*}