# What is the antiderivative of e^(2x)?

Mar 27, 2015

Antiderivative is another name for the Integral( if by some misfortune you didnt know)
So,

$\int {e}^{2 x} = 1 \frac{l}{2} \int 2 {e}^{2 x} \mathrm{dx}$

You can see that $2 \mathrm{dx} = d \left(2 x\right)$

that is $2$ is the derivative of $2 x$

It follows : $\frac{1}{2} \int {e}^{2 x} d \left(2 x\right)$
NOTE: this is the same as letting $u = 2 x$

$\frac{1}{2} \int {e}^{u} \mathrm{du} = \frac{1}{2} {e}^{u}$
$= \frac{1}{2} {e}^{2 x}$

Generally, $\int {e}^{a x} = \frac{1}{a} {e}^{a x}$

Mar 27, 2015

It is $\frac{1}{2} {e}^{2 x}$.

You can certainly use the technique of integration by substitution (reversing the chain rule) to find this, you can also reason as follows:

The antiderivative of ${e}^{2 x}$ is a function whose derivative is ${e}^{2 x}$.

But we know some things about derivatives at this point of the course. Among other things, we know that the derivative of $e$ to a power is $e$ to the power times the derivative of the power.

So we know that the drivative of ${e}^{2 x}$ is ${e}^{2 x} \cdot 2$. That's twice a big as what we want.

We also know that constant factors just hang out in front when we take derivatives, so if we stick a $\frac{1}{2}$ out front, it will be there after we differentiate and we can cancel the two.

$f \left(x\right) = \frac{1}{2} {e}^{2 x}$ has $f ' \left(x\right) = {e}^{2 x}$ so it is an antiderivative. The general antiderivative then is $\frac{1}{2} {e}^{2 x} + C$

Note
An important consequence of the Mean Value Theorem is that a function whose derivative is $0$ is a constant function. And an immediate consequence of that is that if two functions have the same derivative, then they differ by a constant.
Therefore, any function that has derivative ${e}^{2 x}$ can ultimately be written as $\frac{1}{2} {e}^{2 x} + C$ for some constant C.