Video Transcript
Determine the integral of two to
the power of nine π₯ with respect to π₯.
Letβs begin by quoting what we do
know about the integral of π to the power of π₯. Itβs π to the power of π₯ divided
by the natural log of π. Our integrand is slightly different
though; itβs a constant to the power of another constant times π₯. So, weβre going to use the process
of introducing something new, a new letter. We let π’ be equal to nine π₯, and
of course this is known as integration by substitution. We obtain the derivative of dπ’
with respect to π₯ to be equal to nine. Now, remember, dπ’ by dπ₯ is
absolutely not a fraction, but we can treat it a little like one for the purposes of
integration by substitution. And we see that we can say that a
ninth dπ’ equals dπ₯.
We replace π’ with nine π₯ and dπ₯
with a ninth dπ’. And then, we take out this constant
factor of a ninth. And we see that our integral is now
a ninth of the integral of two to the power of π’ with respect to π’. Well, the integral of two to the
power of π’ is two to the power of π’ over the natural log of two. And then, of course, we can use the
definition of our substitution and replace π’ with nine π₯. And weβve found the integral of two
to the power of nine π₯ with respect to π₯. Itβs two to the power of nine π₯
over nine times the natural log of two plus this constant of integration πΆ, which
Iβve made a capital πΆ to show that itβs different from the value we had before.
