Recall that we previously established the following results when differentiating exponential functions:
| $\frac{d}{dx}e^x$ddxex | $=$= | $e^x$ex |
| $\frac{d}{dx}e^{ax+b}$ddxeax+b | $=$= | $ae^{ax+b}$aeax+b |
Reversing these we get the following rules for integrating exponential functions:
$\int e^xdx=e^x+c$∫exdx=ex+c, for some constant $c$c
$\int e^{ax+b}dx=\frac{1}{a}e^{ax+b}+c$∫eax+bdx=1aeax+b+c, for some constant $c$c
Worked examples
Example 1
Determine $\int e^{2-7x}dx$∫e2−7xdx.
Think: To integrate we need to divide by $a$a from the term $ax+b$ax+b; here $a=-7$a=−7. The power of $e$e remains unchanged.
Do:
$\int e^{2-7x}dx=\frac{-1}{7}e^{2-7x}+c$∫e2−7xdx=−17e2−7x+c, where $c$c is a constant.
Example 2
Determine $\int\frac{e^{2x}-3}{e^x}dx$∫e2x−3exdx.
Think: We can split this fraction and rewrite the terms as powers of $e$e. Then we can apply our rule term by term.
Do:
| $\frac{e^{2x}-3}{e^x}$e2x−3ex | $=$= | $\frac{e^{2x}}{e^x}-\frac{3}{e^x}$e2xex−3ex |
Split the fraction |
| $=$= | $e^{2x-x}-3e^{-x}$e2x−x−3e−x |
Use index laws to rewrite |
|
| $=$= | $e^x-3e^{-x}$ex−3e−x |
Simplify |
Hence,
| $\int\frac{e^{2x}-3}{e^x}dx$∫e2x−3exdx | $=$= | $\int e^x-3e^{-x}dx$∫ex−3e−xdx |
| $=$= | $e^x+3e^{-x}+c$ex+3e−x+c, where $c$c is a constant |
Example 3
If $f'\left(x\right)=1+8e^{-2x}$f′(x)=1+8e−2x and $f\left(0\right)=2$f(0)=2, find $f\left(x\right)$f(x).
Think: We can first find the indefinite integral, and then use the given point $\left(0,2\right)$(0,2) to find the value of the constant of integration $c$c.
Do:
| $\int1+8e^{-2x}dx$∫1+8e−2xdx | $=$= | $x-4e^{-2x}+c$x−4e−2x+c |
Using the point $\left(0,2\right)$(0,2), find $c$c:
| $f\left(0\right)$f(0) | $=$= | $2$2 |
| $\therefore\ 0-4e^0+c$∴ 0−4e0+c | $=$= | $2$2 |
| $-4+c$−4+c | $=$= | $2$2 |
| $c$c | $=$= | $6$6 |
Thus, $f\left(x\right)=x-4e^{-2x}+6$f(x)=x−4e−2x+6.
Practice questions
Question 1
Determine $\int4e^{1-2x}dx$∫4e1−2xdx.
You may use $C$C as a constant.
Question 2
A curve has gradient function $\frac{dy}{dx}=e^{3x}$dydx=e3x. Find the equation of the curve if it passes through $\left(0,-1\right)$(0,−1).
Question 3
Given $\frac{dy}{dx}=ke^x+2$dydx=kex+2, for some constant $k$k, and that when $x=0$x=0 we have $y=-1$y=−1 and $\frac{dy}{dx}=5$dydx=5:
Determine the value of $k$k.
Find $\int\left(ke^x+2\right)dx$∫(kex+2)dx using the value of $k$k from part (a). Use $C$C as the unknown constant.
Find $y$y in terms of $x$x.