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IN3.3 Integration of exponential functions


Example of indefinite and definite integral of an exponential function.

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 \[\begin{align*} \int e^{ax+b}dx & =\int\frac{1}{a}e^{u}\frac{du}{dx}dx\\ & =\frac{1}{a}\int e^{u}du\\ & =\frac{1}{a}e^{u}+c\\ & =\frac{1}{a}e^{ax+b}+c. \end{align*}\] \[\begin{align*} \int e^{ax+b}dx & =\frac{1}{a}e^{ax+b}+c,\textrm{ where $a,b$ and $c$ are constants}. \end{align*}\]


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


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


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

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

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

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