 Let's take another look at u-substitutions so we can expand our ability to integrate functions. So remember, our fundamental theorem of calculus says that the definite integral from a to b of f of t dt is capital F of b, the endpoint, minus capital F of a, the beginning point, where capital F of x is any antiderivative of f of x. And a bit of notation. We can express this as the definite integral from x equals a to x equals b, f of x dx, and the evaluation, this f of b minus f of a, we can express as f of x bar, x equals a to x equals b. Well, this is how we can express it. We often drop out the x equals, and so we see this form. However, here's an important idea. When we see this form, there is an implied x equals, x equals a, x equals b, and if in doubt, write it out, paper is cheap. It's sometimes useful to implement the following variation on a u-substitution. So as before, we'll let u be some function of x. We'll find du equals g prime of x dx, and rather than solving for dx, we'll manipulate the integrand so it includes this g prime of x dx. Let's see what that looks like. Suppose I want to find the integral of x square root x squared plus 5 dx. So I'll let u equals x squared plus 5, and so du is 2x dx. Now, when we started out with u-substitutions, we might have solved for dx and then substituted it into our integrand. This time, let's see if we can get a 2x dx in our actual integrand. So now, outside of this square root x squared plus 5, which is going to become part of our u-substitution, we have this x. And so, first of all, let's make that a 2x by multiplying by 2, but then we also have to multiply by 1 half to keep working with the same function. Now, I'll move the 2x over next to the dx, and remember the integral of constant times function is constant times integral. So we can move this 1 half out front of the integral sign. I'll make my u-substitution u equals x squared plus 5, so this square root becomes square root of u, and du is this 2x dx factor, and so that gets replaced. I'll do the integration, put everything back where we found it, and because this is an indefinite integral, remember that we have that added constant of anti-differentiation. And this leads to another important idea. An indefinite integral must be expressed in terms of its original variables. If you're given an integral in x, your answer must also be in x. But if we have a definite integral where we have an upper and lower bound, we can re-parameterize in terms of the new variables. What does that mean? Well, suppose I take the same integrand, x, square root of x squared plus 5, but this time I'm going to find the definite integral from 3 to 5. On the one hand, I can use the anti-derivative I found, and so I'm going to evaluate this function at 5 and subtract this function evaluated at 3, which gives me the value of the definite integral. We can try a different approach. Remember that we often omit the variable when expressing the limits of integration. But if in doubt, write it out. So this integral from 3 to 5 is actually the integral from x equals 3 to x equals 5. Why does that matter? Well, since we made the substitution u equals x squared plus 5, then when x is equal to 3, then u is equal to 14, and when x is equal to 5, u is equal to 30. And that means when I make my u substitution, I can also substitute in the limits of integration, which means that when I get here, I have a choice. I can either substitute u equals x squared plus 5 back in and use the original limits of integration, which are in terms of x, or I can use the fact that I know that u goes from 14 to 30. And so I don't need to do that substitution back. I can use my new limits of integration and evaluate without having to substitute back.