Learning Lab

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: 11 This form may be derived using integration by substitution. Let \(u=ax+b\) then \(du/dx=a.\)Using substitution \[\begin{align*} \int e^{ax+b}dx & =\int\frac{1}{a}e^{u}\frac{du}{dx}dx\\ & =\frac{1}{a}\int e^{u}du\\ & =\frac{1}{a}e^{u}+c\\ & =\frac{1}{a}e^{ax+b}+c. \end{align*}\] \[\begin{align*} \int e^{ax+b}dx & =\frac{1}{a}e^{ax+b}+c,\textrm{ where $a,b$ and $c$ are constants}. \end{align*}\]

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

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:
   \(\begin{aligned}a) & \intop_{0}^{2}e^{3x}dx & b) & \int_{-1}^{3}e^{2-5x}dx & c) & \int_{-1}^{1}\frac{9e^{3x}+5}{e^{2x}}\end{aligned} dx\)