Video: Finding the Integration of an Exponential Function with an Integer Base

Determine ∫ 2^(9π‘₯) dπ‘₯.

01:22

Video Transcript

Determine the integral of two to the power of nine π‘₯ with respect to π‘₯.

Let’s begin by quoting what we do know about the integral of π‘Ž to the power of π‘₯. It’s π‘Ž to the power of π‘₯ divided by the natural log of π‘Ž. Our integrand is slightly different though; it’s a constant to the power of another constant times π‘₯. So, we’re going to use the process of introducing something new, a new letter. We let 𝑒 be equal to nine π‘₯, and of course this is known as integration by substitution. We obtain the derivative of d𝑒 with respect to π‘₯ to be equal to nine. Now, remember, d𝑒 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 d𝑒 equals dπ‘₯.

We replace 𝑒 with nine π‘₯ and dπ‘₯ with a ninth d𝑒. 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 𝑒 with respect to 𝑒. Well, the integral of two to the power of 𝑒 is two to the power of 𝑒 over the natural log of two. And then, of course, we can use the definition of our substitution and replace 𝑒 with nine π‘₯. And we’ve found the integral of two to the power of nine π‘₯ with respect to π‘₯. It’s two to the power of nine π‘₯ over nine times the natural log of two plus this constant of integration 𝐢, which I’ve made a capital 𝐢 to show that it’s different from the value we had before.

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