Integrating Exponentials

We can recall that:

\[ \textrm{If }y=e^{ax}\textrm{ then }\frac{dy}{dx} = a \time e^{ax} \]

So when we are differentiating \(e^{ax}\), we multiply the exponential by the coefficient of the power. When integrating we do the opposite, so we divide by the coefficient in the power. In general, we have:

\[ \int e^{ax}\;dx = \frac{e^{ax}}{a} + C \]

Example

Find the following:

1. \(\int e^{6x}\;dx\)

2. \(\int (e^x + 4e^{-3x})\;dx\)

3. \(\int 9e^{3z}\;dz\)

Solution:

1. Using the rule \(\int e^{ax}\;dx=\frac{e^{ax}}{a} + C\) and dividing by the coefficient of the power, we find:

   \[ \int e^{6x}\;dx = \frac{e^{6x}}{6} + C \]

2. First, we should break up the integral across the sum:

   \[ \int(e^{x} + 4e^{-3x})\;dx = \int e^x\;dx + 4\in e^{-3x}\;dx \]

   Now we can apply the rule above to get:

   \[ \int e^x\;dx + 4\in e^{-3x}\;dx = \frac{e^x}{1} + 4\times \frac{e^{-3x}}{-3} + C = e^x - \frac{4e^{-3x}}{3} + C \]

3. Using the rule above, but this time, with respect to \(z\), we find:

   \[ \int 9e^{3z}\;dz = 9 \int e^{3z}\;dz = 9\times \left(\frac{e^{3z}}{3} + C)\right) + 3e^{3z} + C' \]

   where \(C' = 9C\).

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sympy is well capable of handling the integration of exponentials.

from sympy import symbols, exp, integrate


x = symbols('x')

integrate(exp(6 * x))

\[\displaystyle \frac{e^{6 x}}{6}\]

integrate(exp(x) + 4 * exp(-3 * x))

\[\displaystyle e^{x} - \frac{4 e^{- 3 x}}{3}\]

z = symbols('z')

integrate(9 * exp(3 * z))

\[\displaystyle 3 e^{3 z}\]

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