# IN3.3 Integration of exponential functions

Exponential growth functions | Exponential and logarithmic functions | Algebra II | Khan Academy
Exponential growth functions | Exponential and logarithmic functions | Algebra II | Khan Academy

## IN3.3 Integration of exponential functions

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

$\begin{array}{llllllllll} a) & \frac{e^{3x}}{3}+c & & & b) & -\frac{e^{2-5x}}{5}+c & & & c) & 9e^{x}-\frac{5}{2e^{2x}}+c\end{array}$

$\begin{array}{llllllllll} a) & \frac{e^{6}}{3}-\frac{1}{3} & & & b) & \frac{1}{5}\left(e^{7}-e^{-13}\right) & & & c) & 9\left(e-e^{-1}\right)-\frac{5}{2}\left(e^{2}-e^{-2}\right)\end{array}$