Earlier in the course, we learned how to evaluate indefinite integrals using the process of substitution. And now, we'll need to use substitution to evaluate definite integrals as well. Luckily, we'll be able to use many of the same steps that we know from working with our indefinite integrals. The only thing that's different is our integral now has bounds. It turns out that there are two different ways that we could choose to address the bounds of our definite integral.
And I'm going to walk you through both of those methods here. So let's go ahead and get started by jumping right into our example. We're asked to evaluate the definite integral using substitution. And we're given the integral from zero to two of x2+13×2xdx. Now, we've worked with this exact integral before.
The only difference is it's now bounded from zero to two. So how exactly do we address those bounds, and where does that fit into our steps? Well, for this first method, we're actually going to start by completely ignoring our bounds and treating this as though it's just an indefinite integral. So getting right into our first couple of steps here, we know that we need to choose u and find du so that we can rewrite our integral just in terms of them.
Now, having worked with this exact integral before, we know that if we choose u to be x2+1, that then makes du≡2x dx, allowing us to rewrite our integral as the integral of u3du. Now having completed those first two steps here, we can move on to step number three where we want to go ahead and integrate here with respect to u. Now getting our antiderivative, I know that this is going to give me 14u4 using the power rule. Then I need to add that constant of integration c because this is currently an indefinite integral. Now with that third step done, we move on to step four, which is replacing u with that original function of x.
In this case, x2+14. So plugging that x2+1 in for u here, we get 14 times x2+14, and then plus that constant of integration c. Now if this were actually an indefinite integral, we would be completely done at this point. But remember, this is not an indefinite integral and it is in fact bounded from zero to two. So here's where our bounds come in.
I'm going to go ahead and get out of the way as we add an additional step to our steps box here in order to address those bounds. At this point, we want to go ahead and evaluate the antiderivative that we found at those original bounds. In this case, zero and two. So looking at our antiderivative here, that 14 x2+14, I don't need to worry about that constant of integration c here because this is a definite integral, and I want to go ahead and bound it from zero to two. Now we can use the fundamental theorem of calculus here to plug in those bounds and get our final answer.
So this is going to be one fourth times two squared plus one to the power of four minus zero squared plus one to the power of four, plugging in my upper and lower bounds. Now when I work this out algebraically I will end up getting a final answer here of 156 having evaluated this definite integral using substitution. Now for this first method here, we wanted to treat this as an indefinite integral, doing our substitution the way that we know how, putting our function back in terms of x, and then applying those original bounds. But remember that there's another way that we could choose to address our bounds. So let's take a look at our second method here.
Now we're faced with the same exact definite integral here from zero to two of x2+13×2xdx. Now, having worked with this integral so many times before, we already know that if we choose u as x2+1 that makes du≡2x dx, allowing us to rewrite our integral here as the integral of u3du. Now at this point in our previous method, we were completely ignoring the bounds of our integral, but that's not what we're going to do here. Instead, we're going to transform our bounds along with the rest of our integral. So having completed step one already here, as we are rewriting our integral only in terms of u and du, we're going to add one additional part to this step here where we're going to address our bounds.
So we want to go ahead and transform our bounds using the substitution that we made, plugging our original bounds into that function of x. So whereas our original bounds were from a to b, as we rewrite our new integral in terms of u and we have that f(u)du, this will now be bounded from g(a) to g(b), having plugged that a and b into that function of x. In this case, since we chose u as x2+1, I want to plug zero and two into that. So g(0) here, plugging that in, is going to be zero squared plus one, which is just one. And then g(2), plugging that two in, that's two squared plus one, which gives me five.
So this one and five represent my brand new transformed bounds. So this integral of u3 du is bounded from one to five. Now we have fully rewritten our entire integral in terms of u and du, including those bounds. And we can go ahead and now move on to step three. We want to integrate here with respect to u as we've done before.
Now having worked with this so many times, we already know that this will come out to 14 u4, and we don't have to worry about our constant of integration because we have bounds. This is bounded from one to five. And since we have those brand new bounds, moving on to that fourth step here, we no longer need to put this back in terms of x. So we're going to completely get rid of that step here and instead add one final step to finish evaluating this integral by going ahead and evaluating this antiderivative at these brand new bounds that we found. So going ahead and plugging these in here using the fundamental theorem of calculus, this is going to be one fourth times five to the power of four minus one to the power of four.
Now this will work out algebraically to give us an answer of 156 for this definite integral, which is the same answer that we got up above in method one. Now between these two methods, there is not one best method, and they'll both always work. But no matter what method you choose to use, make sure you know exactly how to apply your bounds. In that first method, we treat it as an indefinite integral, doing our substitution, putting it back in terms of x, and applying those original bounds. In our second method here, we put everything in terms of u when doing our substitution so that we could keep it in terms of u and evaluate our antiderivative at those brand new transformed bounds.
Now we're going to continue getting practice with substitution coming up next. I'll see you there.
