Chapter 3 Section 7

3.7 Definite Integral of the Exponential Function

Preliminary Results

We will use the following results.

• The Laws of Exponents. Particularly, \(a^ha^k=a^{h+k}\), and \((a^h)^k= a^{hk}\) for \(a >0 \).

 

• Continuity of the exponential function

 

• The limit of the difference quotient of \(e^x\):
  \begin{equation}\label{ch03-07-eq01}
  \lim_{x\to 0}\frac{e^x-1}{x} = 1\tag{3.11}.
  \end{equation}

 

• The factorization formula
  \[a^n – b^n = (a – b)\left(a^{n-1} + a^{n-2}b + a^{n-3} b^2+ \cdots + a^2b^{n-3}+ab^{n-2}+b^{n-1}\right).\]

  Actually, we will need this formula for the case \(a=1\) and \(b\neq 1\), in the following form:
  \begin{equation}\label{ch03-07-eq02}
  b\frac{b^n-1 }{b-1} =b + b^2 + b^3+ \cdots + b^{n-2}+b^{n-1}+b^n =\sum _{i=1}^nb^i\tag{3.12}.
  \end{equation}
  Such a sum is called a \textbf{geometric sum}.

3.7.1 Definite Integral of the Exponential Function

Proposition 1. Given \(a,\, b \in \mathbb{R}\) with \(a < b\), then \begin{equation}\label{ch03-07-eq03} \int_a^be^x\,dx = e^b-e^a\tag{3.13}. \end{equation}

Proof.

Step 1. Partition of the interval.
\[\Delta x= \frac{b-a}{n}, \;\; x_i=a+i \Delta x=a+\frac{(b-a)i}{n}, \;\; I_i=[x_{i-1},x_i].\]
 

Step 2. Height and area of the rectangular regions.
\[h_i=e^{x_i}=e^{a+i\Delta x} = e^a e^{i\Delta x}
= e^a \left(e^{\Delta x}\right)^i, \;\;\;A_i = e^{x_i}\Delta x =e^a \left(e^{\Delta x}\right)^i\Delta x.\]
 

Step 3. Computation of \(R(e^x,n)\).
Here we are going to use (\ref{ch03-07-eq02}) with \(b=\left(e^{\Delta x}\right)\).
\begin{align*}
R(e^x,n)&=\sum _{i=1}^n e^{x_i}\Delta x
=\sum _{i=1}^n e^a \left(e^{\Delta x}\right)^i \Delta x
=e^a\Delta x\sum _{i=1}^n \left(e^{\Delta x}\right)^i
\underset{(1)}{=}
e^a\Delta x e^{\Delta x}\frac{\left(e^{\Delta x}\right)^n-1}{e^{\Delta x}-1} =\\
&\underset{(?)}{=}e^a\Delta x \frac{e^{\Delta x}}{e^{\Delta x}-1}
\left(\left(e^{\Delta x}\right)^n-1\right)
\underset{(2)}{=}e^a\Delta x \frac{e^{\Delta x}}{e^{\Delta x}-1}\left(e^{(b-a)}-1\right)
\underset{(3)}{=} e^{\Delta x}\frac{\Delta x}{e^{\Delta x}-1}(e^b-e^a)
.
\end{align*}
(1) Here is where we used (\ref{ch03-07-eq02}). (2) Replace \(\left(e^{\Delta x}\right)^n = \left(e^{\frac{b-a}{n}}\right)^n=e^{b-a}\). (3) Distribute the factor \(e^a\).

Step 4. Evaluation of the limit of \(\displaystyle \lim_{n\to \infty} R(e^x,n)\).

Given that \(\displaystyle \Delta x =\frac{b-a}{n}\), with \(b-a\) fixed, then
\[\Delta x \to 0 \Longleftrightarrow n\to \infty,\]
that is, increasing the number of intervals in the partition is equivalent to letting \(\Delta x \to 0\). Thus
\[\lim_{n\to \infty}\sum _{i=1}^n e^{x_i}\Delta x
=\lim_{\Delta x \to 0} e^{\Delta x}\frac{\Delta x}{e^{\Delta x}-1}(e^b-e^a)
\underset{(\ref{ch03-07-eq01})}{=} e^0(1)(e^b-e^a) = e^b-e^a.\]

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