Integration of exponential functions

How do you integrate an exponential function? This skill is important for calculating compound interest over time in finance, determining population growth models in biology, and analysing radioactive decay in physics. Use this resource to learn how.

Exponential functions have the form:
\[y=ae^{kx}\]

where \(a\) and \(k\) are constants. Examples include \(y=e^{2x+3}\), \(f(x)=e^{3x}\) and \(y=e^{x-1}\).

The antiderivative or indefinite integral of \(e^{x}\) is \(e^{x}\). Therefore:

\[\int e^{x}dx=e^{x}+c\]

where \(c\) is a constant.

More generally:

\[\int e^{ax+b}dx = \frac{1}{a}e^{ax+b}+c\]

where \(a\), \(b\) and \(c\) are constants.

Example 1 – integrating exponential functions

Integrate \(2e^{x}\) with respect to \(x\).
\[\int2e^{x} = 2e^{x}+c\]

Find \(\int e^{-5x+1}dx\).

Here, \(a=5\) and \(b=1\).
\[\int e^{-5x+1}dx = -\frac{1}{5}e^{-5x+1}+c\]

Find the integral \(\int_{1}^{2}2e^{x}dx\).

The limits of the integral are \(1\) and \(2\). We have:
\[\begin{align*} \int_{1}^{2}2e^{x}dx & = \left[2e^{x}+c\right]_{x=1}^{x=2}\\
& = \left(2e^{2}+c\right)-\left(2e^{1}+c\right)\\
& = 2e^{2}+c-2e^{1}-c\\
& = 2e^{2}-2e^{1}
\end{align*}\]

Evaluate \(\int_{-1}^{4}2e^{x}dx\).
\[\begin{align*} \int_{-1}^{\,4}2e^{x}dx & = \left[2e^{x}\right]_{x=-1}^{x=4}\\
& = 2e^{4}-2e^{-1}
\end{align*}\]

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

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

Exercise – integrating exponential functions

Calculate the following.

\(\int e^{3x}dx\)
\(\int e^{2-5x}dx\)
\(\int \dfrac{9e^{3x}+5}{e^{2x}}dx\)
Hint: Divide through first.

Evaluate the following.

\(\int_{0}^{2}e^{3x}dx\)
\(\int_{-1}^{3}e^{2-5x}dx\)
\(\int_{-1}^{1}\dfrac{9e^{3x}+5}{e^{2x}}dx\)

1. \(\dfrac{e^{3x}}{3}+c\)
2. \(-\dfrac{e^{2-5x}}{5}+c\)
3. \(9e^{x}-\dfrac{5}{2e^{2x}}+c\)
1. \(\dfrac{e^{6}}{3}-\dfrac{1}{3}\)
2. \(\dfrac{1}{5}(e^{7}-e^{-13})\)
3. \(9(e-e^{-1})+\dfrac{5}{2}(e^{2}-e^{-2})\)