IN3.3: Integration of exponential functions

The integration of exponential functions is widely used in various fields such as physics, engineering, and economics to model and solve problems involving growth, decay, and other processes that change exponentially over time. Keep reading to find out more.

This module deals with differentiation of exponential functions such as: \[\begin{align*} & \int\exp\left(2x+3\right)dx\\ & \int e^{3x}dx\\ & \int_{1}^{2}e^{x-1}dx. \end{align*}\]

Indefinite integral of an exponential function

If \(f\left(x\right)=e^{x}\) then \(f'\left(x\right)=e^{x}.\) Therefore an antiderivative (or indefinite integral) of \(e^{x}\) is \(e^{x}\). That is \[\begin{align*} \int e^{x}dx & =e^{x}+c,\textrm{ where $c$ is a constant.} \end{align*}\] A more general form is: 1 This form may be derived using integration by substitution. Let \(u=ax+b\) then \(du/dx=a.\)Using substitution \[\begin{align*} \int e^{ax+b}dx & =\int\frac{1}{a}e^{u}\frac{du}{dx}dx\\ & =\frac{1}{a}\int e^{u}du\\ & =\frac{1}{a}e^{u}+c\\ & =\frac{1}{a}e^{ax+b}+c. \end{align*}\] \[\begin{align*} \int e^{ax+b}dx & =\frac{1}{a}e^{ax+b}+c,\textrm{ where $a,b$ and $c$ are constants}. \end{align*}\]

Examples

1. \(\int2e^{x}dx=2e^{x}+c\).
2. \(\int e^{-5x+1}dx=-\frac{1}{5}e^{-5x+1}+c\), \(\left(a=5,\;b=1\right)\).
3. \(\int e^{\frac{x}{3}+4}dx=\frac{1}{1/3}e^{\frac{x}{s}+4}+c=3e^{\frac{x}{3}+4}+c\), \(\left(a=\frac{1}{3},\;b=4\right)\).

Definite integral of an exponential function

Now that we know how to get an antiderivative (or indefinite integral) of an exponential function we can consider definite integrals. To evaluate a definite integral we determine an antiderivative and calculate the difference of the values of the antiderivative at the limits defined in the definite integral. For example consider

\[\begin{align*} \int_{1}^{2}2e^{x}dx. \end{align*}\] From the previous section we know an antiderivative is \(2e^{x}+c\) where \(c\) is a constant. The limits of the integral are \(1\) and \(2\). So we have \[\begin{align*} \int_{1}^{2}2e^{x}dx & =\left[2e^{x}+c\right]_{x=1}^{x=2} & \left(1\right)\\ & =\left(2e^{2}+c\right)-\left(2e^{1}+c\right) & \left(2\right)\\ & =2e^{2}+c-2e^{1}-c & \left(3\right)\\ & =2e^{2}-2e^{1}. & \left(4\right) \end{align*}\]

Note that the notation in line \(\left(1\right)\) \[\begin{align*} \left[2e^{x}+c\right]_{x=1}^{x=2} \end{align*}\] means substitute \(x=2\) in the expression in brackets and subtract the expression in brackets evaluated at \(x=1.\)

Note also that the constant \(c\) in lines \(\left(1\right)\) to \(\left(3\right)\) has no effect when evaluating a definite integral. Consequently we usually leave it out and write \[\begin{align*} \int_{1}^{2}2e^{x}dx & =\left[2e^{x}\right]_{x=1}^{x=2}\\ & =\left(2e^{2}\right)-\left(2e^{1}\right)\\ & =2e^{2}-2e^{1}. \end{align*}\]

Examples

1. Evaluate \(\int_{-1}^{\,4}2e^{x}dx.\)
   Solution: \[\begin{align*} \int_{-1}^{\,4}2e^{x}dx & =\left[2e^{x}\right]_{x=-1}^{x=4}\\ & =2e^{4}-2e^{-1}. \end{align*}\]
2. Evaluate \(\int_{0}^{2}e^{-5x+1}dx\).
   Solution: \[\begin{align*} \int_{0}^{2}e^{-5x+1}dx & =\left[-\frac{1}{5}e^{-5x+1}\right]_{x=0}^{x=2}\\ & =\left(-\frac{1}{5}e^{-5\left(2\right)+1}\right)-\left(-\frac{1}{5}e^{-5\left(0\right)+1}\right)\\ & =\left(-\frac{1}{5}e^{-9}\right)-\left(-\frac{1}{5}e^{1}\right)\\ & =-\frac{1}{5}e^{-9}+\frac{1}{5}e\\ & =\frac{1}{5}\left(e-e^{-9}\right). \end{align*}\]
3. Evaluate \(\int_{-3}^{\,9}e^{\frac{x}{3}+4}dx\).
   Solution: \[\begin{align*} \int_{-3}^{\,9}e^{\frac{x}{3}+4}dx & =\left[3e^{\frac{x}{3}+4}\right]_{x=-3}^{x=9}\\ & =\left(3e^{\frac{9}{3}+4}\right)-\left(3e^{\frac{-3}{3}+4}\right)\\ & =\left(3e^{3+4}\right)-\left(3e^{-1+4}\right)\\ & =3\left(e^{7}-e^{3}\right). \end{align*}\]

Exercises

1. Calculate:
   \(\begin{aligned}a) & \intop e^{3x}dx & b) & \int e^{2-5x}dx & c) & \int\frac{9e^{3x}+5}{e^{2x}}dx\text{ Hint: Divide through first.}\end{aligned}\)
2. Evaluate:
   \(\begin{aligned}a) & \intop_{0}^{2}e^{3x}dx & b) & \int_{-1}^{3}e^{2-5x}dx & c) & \int_{-1}^{1}\frac{9e^{3x}+5}{e^{2x}}\end{aligned} dx\)