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

2 Answers
Mar 27, 2015

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

#inte^(2x) = 1l/2int 2e^(2x) dx#

You can see that #2dx = d(2x)#

that is #2# is the derivative of #2x#

It follows : #1/2int e^(2x) d(2x)#
NOTE: this is the same as letting #u = 2x#

#1/2int e^u du = 1/2e^u #
# = 1/2e^(2x)#

Generally, #int e^(ax) = 1/ae^(ax)#

Mar 27, 2015

It is #1/2 e^(2x)#.

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^(2x)# is a function whose derivative is #e^(2x)#.

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^(2x)# is #e^(2x)*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 #1/2# out front, it will be there after we differentiate and we can cancel the two.

#f(x)=1/2e^(2x)# has #f'(x)=e^(2x)# so it is an antiderivative. The general antiderivative then is #1/2 e^(2x) +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^(2x)# can ultimately be written as #1/2 e^(2x)+C# for some constant C.