Find the indefinite integral.∫r2e−rdr\int_{}^{}r^2e^{-r}dr∫r2e−rdr

− r 2 e − r − 2 r e − r − 2 e − r + C -r^2e^{-r}-2re^{-r}-2e^{-r}+C − r 2 e − r − 2 r e − r − 2 e − r + C

Find the indefinite integral. ∫ r 2 e − r d r \int_{}^{}r^2e^{-r}dr ∫ r 2 e − r d r

in this problem, we're asked to find the indefinite integral of r2 times e to the negative r with respect to r. So to solve this problem, what you may notice is we have this situation where we have a product of two functions, but it turns out there's not really a place we can do a variable substitution. Doing a u substitution on the r here is not going to get rid of this r that we have, that e is raised to. So we have e to the negative r, that's not gonna clear that, And if I do a u substitution up here, we're not going to be able to get rid of this r squared out front, so we need another strategy. So since this method is failing, we're going to try integration by parts and see if that's going to allow us to get a solution. Now recall we know the general formula for integration by parts. The integral of u d v is equal to u v minus the integral of v d u. Now what I first need to do with this formula is figure out what my u is. Now recall u is gonna be whatever becomes easier as you take the derivative. Now looking at my integral here, I can see we have this r squared. This is the algebraic term that I'm looking at, and I know that r squared will become more simple if I take its derivative. It'll become two r. So I'm gonna go ahead and say that r squared is u. I mean, if I take the derivative of u here with respect to r, well, that's going to, using the power rule, give me two r, and that's gonna be with respect to r. Notice how we were able to get from an r squared to just an r by itself, except multiplied by two here, obviously, but I mean, it went from r squared to not having any kind of exponent on it. So we can say that that did get more simple, meaning that was the correct u that we selected. Now next, I need to find my dv. Now to find my dv, this should be something that's easily integrated. Now I see we have e to a power of negative r. I know these e functions in general are relatively straightforward to integrate, so I'm going to say that e to the negative r with respect to r is my dv. So what I need to do is figure out what the integral of this is. Well integrating dv is going to give me that v is equal to the integral of e to the negative r doctor. If you go ahead and integrate e to the negative r, you're just gonna get negative e to the negative r. You can go ahead and go through this integral right here, and what would end up happening is this negative sign right here would be divided on the outside as negative one, which would just give you this negative e to the negative r. So that's how we can find our u and our du and our dv and our v. So now that we have these, we can go ahead and plug them into this side of the formula, uv minus the integral of vdu. So plugging things in we're going to have u times v, which I can see is gonna be r squared times negative e to the negative r, then it's going to be minus the integral of v du. I'll go ahead and move this down here so we can fit this all in. So it's gonna be minus the integral v du, so we're gonna have minus the integral of this product right here, which is gonna be negative e to the negative r, then we have du, which is two r doctor. Now I'm gonna go ahead and clean this up a little bit. So what I can do is cancel the negative sign right here, and I can move this negative sign out to the front. So it's gonna give me negative r squared e to the negative r, and we're going to have plus, and then I'll take this two out to the front as well. We'll have two, and that's gonna be times the integral of r, and that's gonna be e to the negative r doctor. So this is what we get. Now notice that we still have not gotten to a final solution yet, and if we were to try integrating this, we've run into another issue because we can't do variable substitutions or any other rules here either. But notice we can use integration by parts a second time, and this integration by parts looks like it's already gonna be more simple than what we had up there. Since I can see things are getting more simple, I know that I'm doing this pattern correctly, so let's go ahead and use this formula a second time. So we have this formula right here, and now we need to pick a new u and a new dv. Well going through this process the u is going to be something that becomes more simple with the derivative. Before I had it be r squared so I think it makes sense that my u here would just be this r. Then I can take the derivative of u with respect to r giving me just one doctor. So now we've found u and du and now next I'm going to find my dv. Well dv is whatever can be easily integrated which I know is just going to be this e to the negative r. Now we already figured out that v when we integrate this is going to be negative e to the negative r. So now we found our u du, our d v and v, meaning we can use this side of the equation again specifically on this portion because that's what we're trying to integrate. So writing this all out, we're gonna have negative r squared e to the negative r, then we're gonna have plus two, and then in brackets here, we're going to write out this. So we're gonna have u times v, which in this case is gonna be r times negative e to the negative r, and that's going to be minus the integral of v du, which I can see here is going to be we have v there, so it's gonna be negative e to the negative r, and then d, which is doctor, and then that's going to be how we wrap that up there. And so now we just need to evaluate this integral right here and simplify, and that should give us our answer. So I'll rewrite this all out. So we have negative r squared e to the negative r. That's going to be times or or plus two, I should say, times everything in these brackets, which I can write this as negative r e to the negative r. Then I can cancel these negatives here, giving me plus the integral of e to the negative r which I already know is going to give me negative e to the negative r if I integrate this then that's gonna be plus the cost of integration c, and then from here just to go one extra step I'm gonna take this two that I see and distribute it into each of these brackets to get my final answer. So doing this we're going to have this negative r squared e to the negative r that's going to be minus and then that's gonna be two r e to the negative r since you see this negative sign right here let's make that a minus, Then we're going to have two times this, which would be minus two e to the negative r. Then we're gonna have plus two times c, which I can just write as some big constant c since that's just gonna be another constant. Meaning, this right here is gonna be my result and the solution to the problem. So that is how we can do repeated integration by parts to solve this example right here. You can see that we did have to repeat this a couple of times. We had to use this equation more than once, but we were still able to get to an answer. And as you can see, it was really just about whatever we said to you becoming more and more simple until we finally got to an integral that we could actually deal with right down here. So I hope you found this video helpful. Let me know if you have any questions, and let's move on and get some more practice.

Find the indefinite integral. ∫ r 2 e − r d r \int_{}^{}r^2e^{-r}dr ∫ r 2 e − r d r

− r 2 e − r − 2 r e − r − 2 e − r + C -r^2e^{-r}-2re^{-r}-2e^{-r}+C − r 2 e − r − 2 r e − r − 2 e − r + C

r 2 e − r − 2 r e − r + 2 e − r + C r^2e^{-r}-2re^{-r}+2e^{-r}+C r 2 e − r − 2 r e − r + 2 e − r + C

− 1 2 r 2 e − r − r e − r − e − r + C -\frac12r^2e^{-r}-re^{-r}-e^{-r}+C − 2 1 r 2 e − r − r e − r − e − r + C

1 2 r 2 e − r − r e − r + e − r + C \frac12r^2e^{-r}-re^{-r}+e^{-r}+C 2 1 r 2 e − r − r e − r + e − r + C

11. Techniques of Integration

Integration by Parts

Business Calculus

11. Techniques of Integration

Integration by Parts

Struggling with Business Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the indefinite integral.
$\int_{}^{}r^2e^{-r}dr$

$r^2e^{-r}-2re^{-r}+2e^{-r}+C$

$-r^2e^{-r}-2re^{-r}-2e^{-r}+C$

$-\frac12r^2e^{-r}-re^{-r}-e^{-r}+C$

$\frac12r^2e^{-r}-re^{-r}+e^{-r}+C$

Verified step by step guidance

Step 1: Recognize that the integral involves a product of functions, specifically r^2 and e^(-r). This suggests that integration by parts is a suitable method to solve the problem. Recall the formula for integration by parts: ∫u dv = uv - ∫v du.

Step 2: Choose u = r^2 and dv = e^(-r) dr. This choice is made because differentiating r^2 simplifies it, while integrating e^(-r) is straightforward. Compute du = 2r dr and v = -e^(-r) (since the integral of e^(-r) is -e^(-r)).

Step 3: Apply the integration by parts formula. Substitute u, v, du, and dv into the formula: ∫r^2 e^(-r) dr = -r^2 e^(-r) - ∫(-2r e^(-r)) dr.

Step 4: Simplify the remaining integral, ∫(-2r e^(-r)) dr. Again, use integration by parts for this term. Let u = r and dv = -2e^(-r) dr. Then, du = dr and v = 2e^(-r). Substitute into the formula: ∫(-2r e^(-r)) dr = -2(r e^(-r) - ∫e^(-r) dr).

Step 5: Simplify further by integrating ∫e^(-r) dr, which equals -e^(-r). Combine all terms, including the constant of integration C, to express the final result. Ensure all terms are simplified and written in the correct format.

8:30m

Watch next

Master Introduction to Integration by Parts with a bite sized video explanation from Patrick

Start learning

Integration by Parts practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice