Determine the integral of two to
the power of nine 𝑥 with respect to 𝑥.
Let’s begin by quoting what we do
know about the integral of 𝑎 to the power of 𝑥. It’s 𝑎 to the power of 𝑥 divided
by the natural log of 𝑎. Our integrand is slightly different
though; it’s a constant to the power of another constant times 𝑥. So, we’re going to use the process
of introducing something new, a new letter. We let 𝑢 be equal to nine 𝑥, and
of course this is known as integration by substitution. We obtain the derivative of d𝑢
with respect to 𝑥 to be equal to nine. Now, remember, d𝑢 by d𝑥 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 d𝑢 equals d𝑥.
We replace 𝑢 with nine 𝑥 and d𝑥
with a ninth d𝑢. 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 𝑢 with respect to 𝑢. Well, the integral of two to the
power of 𝑢 is two to the power of 𝑢 over the natural log of two. And then, of course, we can use the
definition of our substitution and replace 𝑢 with nine 𝑥. And we’ve found the integral of two
to the power of nine 𝑥 with respect to 𝑥. It’s two to the power of nine 𝑥
over nine times the natural log of two plus this constant of integration 𝐶, which
I’ve made a capital 𝐶 to show that it’s different from the value we had before.