The integral of $e^x$ex and $e^{ax+b}$eax+b
We established the following results when differentiating exponential functions:
| $\frac{d}{dx}e^x$ddx​ex | $=$= | $e^x$ex |
| $\frac{d}{dx}e^{ax}$ddx​eax | $=$= | $ae^{ax}$aeax |
Reversing these we get these results 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=1a​eax+b+C, for some constant $C$C
Worked examples
Example 1
Find the indefinite integral $\int e^{2-7x}dx.$∫e2−7xdx.
Solution:
$\int e^{2-7x}dx=\frac{-1}{7}e^{2-7x}+C$∫e2−7xdx=−17​e2−7x+C
Example 2
Find the definite integral $\int_0^25e^xdx$∫20​5exdx.
Solution:
| $\int_0^25e^xdx$∫20​5exdx | $=$= | $5[e^x]_0^2$5[ex]20​ |
|  | $=$= | $5(e^2-e^0)$5(e2−e0) |
|  | $=$= | $5(e^2-1)$5(e2−1) |
Â
Other examples
Here is a table listing a few results found by using this rule:
| $f'(x)$f′(x) | $f(x)$f(x) |
|---|---|
| $e^{-4x}$e−4x | $-\frac{1}{4}e^{-4x}+C$−14​e−4x+C |
| $6e^{2-3x}$6e2−3x | $-2e^{2-3x}+C$−2e2−3x+C |
| $\frac{4}{e^{2x+1}}=4e^{-\left(2x+1\right)}$4e2x+1​=4e−(2x+1) | $\frac{-2}{e^{2x+1}}+C$−2e2x+1​+C |
| $\frac{e^x+e^{-x}}{2}$ex+e−x2​ | $\frac{e^x-e^{-x}}{2}+C$ex−e−x2​+C |
Practice questions
Question 1
Determine $\int4e^{1-2x}dx$∫4e1−2xdx.
You may use $C$C as a constant.
Question 2
Find the exact value of $\int_0^3\left(2e^{5x}+e^{-7x}\right)dx$∫30​(2e5x+e−7x)dx.
Question 3
Find the exact value of $\int_{\ln2}^{\ln6}\frac{e^{3x}+9}{e^x}dx$∫ln6ln2​e3x+9ex​dx.
The integral of $a^x$ax
We established the following result in an earlier chapter:
$\frac{d}{dx}a^x=\ln a\ a^x$ddx​ax=lna ax
Reversing the result we get the integral of $a^x$ax.
$\int a^xdx=\frac{1}{\ln a}a^x+C$∫axdx=1lna​ax+C
Â
Worked example
Example 3
Find the indefinite integral $\int3^xdx$∫3xdx.
Solution:
$\int3^xdx=\frac{1}{\ln3}3^x+C$∫3xdx=1ln3​3x+C
Knowing how to find the indefinite and definite integral means we can find the area under a curve given the function and the boundaries.
Finding $f\left(x\right)$f(x) given $f'\left(x\right)$f′(x)
Since we know how to integrate, we can find the primitive function if we are given the derivative $f'\left(x\right)$f′(x) and at least one point that satisfies the function $f\left(x\right)$f(x). This is particularly useful in a number different contexts, for example science or business.
Worked example
Example 4
If $f'\left(x\right)=1+4e^{-x}$f′(x)=1+4e−x and $f\left(0\right)=2$f(0)=2, find $f\left(x\right)$f(x).
Â
Do: Integrating $f'\left(x\right)=1+4e^{-x}$f′(x)=1+4e−x
| $f\left(x\right)$f(x) | $=$= | $x-4e^{-x}+C$x−4e−x+C |
Substituting $f\left(0\right)=2$f(0)=2:
| $0-4e^0+C$0−4e0+C | $=$= | $2$2 |
| $-4+C$−4+C | $=$= | $2$2 |
| $C$C | $=$= | $6$6 |
| $\therefore\ f\left(x\right)$∴ f(x) | $=$= | $x-4e^{-x}+6$x−4e−x+6 |
Â
Anti-differentiation strategies
There are other instances where the idea of anti-differentiation can be used to established more complex results.
As a simple example, suppose we correctly determine that $\frac{\mathrm{d}}{\mathrm{d}x}e^{x^2}=2xe^{x^2}$ddx​ex2=2xex2.
Then we also know that:
$\int xe^{x^2}dx=\frac{1}{2}\int2xe^{x^2}dx=\frac{1}{2}e^{x^2}+C$∫xex2dx=12​∫2xex2dx=12​ex2+C
Slightly more difficult situations arise where integrals can be determined indirectly. In certain circumstances, a known differentiation result can be treated as an equation that can be manipulated to find an integration result. Carefully consider this next example:Â
Suppose we first establish the following result using the product rule:
 $\frac{\mathrm{d}}{\mathrm{d}x}xe^x=e^x\left(x+1\right)$ddx​xex=ex(x+1)
Then we can determine the integral $\int xe^xdx$∫xexdx with the following strategy:
| $\frac{\mathrm{d}}{\mathrm{d}x}xe^x$ddx​xex | $=$= | $e^x(x+1)$ex(x+1) |
| $\frac{\mathrm{d}}{\mathrm{d}x}xe^x$ddx​xex | $=$= | $xe^x+e^x$xex+ex |
| $\frac{\mathrm{d}}{\mathrm{d}x}xe^x-e^x$ddx​xex−ex | $=$= | $xe^x$xex |
| $\int\left(\frac{\mathrm{d}}{\mathrm{d}x}xe^x\right)dx-\int e^xdx$∫(ddx​xex)dx−∫exdx | $=$= | $\int xe^xdx$∫xexdx |
| $xe^x-e^x$xex−ex | $=$= | $\int xe^xdx$∫xexdx |
| $\therefore\int xe^xdx$∴∫xexdx | $=$= | $xe^x-e^x$xex−ex |
Â
Practice questions
Question 4
Consider the following.
Given that $y=e^{3x}\left(x-\frac{1}{3}\right)$y=e3x(x−13​), determine $y'$y′.
You may use the substitutions $u=e^{3x}$u=e3x and $v=\left(x-\frac{1}{3}\right)$v=(x−13​) in your working.
Hence find the exact value of $\int_6^9xe^{3x}dx$∫96​xe3xdx.
Question 5
Determine the exact area bounded by the curve $y=e^{-6x}$y=e−6x, the $x$x-axis, and the lines $x=-2$x=−2 and $x=1$x=1.
Question 6
The acceleration of a particle is given by $a\left(t\right)=2e^{3t}$a(t)=2e3t m/s2, and its velocity is $10$10 m/s initially.
Find its velocity $v\left(t\right)$v(t), after $t=2$t=2 seconds. Give your answer correct to the nearest metre per second.
Let $x\left(t\right)$x(t) be the displacement of the particle at $t$t seconds. If its initial position is $2$2 m to the left of the origin, find its displacement after $4$4 seconds, correct to the nearest metre.