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

Question

Isabella Ackerman · Accepted Answer

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)#

Emily Ammerman · Answer

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.

Adam Adkins · Answer

undefined The antiderivative of $ e^{2x} $ is $ \frac{1}{2}e^{2x} + C $, where $ C $ is the constant of integration.