Evaluate the indefinite integral.∫(s2+4)e3sds\int_{}^{}\left(s^2+4\right)e^{3s}ds

e 3 s 3 ( s 2 + 4 ) − 2 9 s e 3 s + 2 27 e 3 s + C \frac{e^{3s}}{3}\left(s^2+4\right)-\frac29se^{3s}+\frac{2}{27}e^{3s}+C

Evaluate the indefinite integral. ∫ ( s 2 + 4 ) e 3 s d s \int_{}^{}\left(s^2+4\right)e^{3s}ds

So in this problem, we are asked to evaluate the indefinite integral. Now notice here that I have these two functions being multiplied. Now when that happens, a good place to start is to try a variable substitution somewhere. Now Now I'm just gonna tell you now, if you try a variable substitution right here, you're going to find that as you take the derivative and work through the problem, it's not going to get rid of this three s in the raised to this exponent of e. If you try a variable substitution right there, it's not gonna clear this s here, so we're gonna be stuck. So what I'm going to do is try the integration by parts strategy to solve this. Now we've learned a couple different ways to deal with integration by parts. Over this problem, I'm going to try using a table, so that way I can organize all my terms well. So So recall that when using a table, you're going to have two columns. You're going to have the derivative, and then you're going to have the integral. Now the derivative is equivalent to what you set u equal to when doing integration by parts, and the integral is what you set d v equal to. So I need to find my u and d v in this problem. Now notice right up front here, we have this polynomial, s squared plus four, being multiplied by e to the three s. I think that's gonna be a good thing to set equal to my u, Because u needs to become more simple as I take its derivative. And I know if I take the derivative of these polynomials, it will become more simple. Whereas if I take the derivative of e to the three s, well, e functions tend to just kinda repeat when you take its derivative or integrate. So I think that this is going to be a better u to pick. So I'm gonna say that s squared plus four is going to be my u. Now that means that my d v is going to be what's ever easily integrated, which in this place I'm going to make it this e to the three s right here. So e to the three s is going to be in my I column. Now what I need to do is keep taking derivatives and keep taking integrals, but I'm gonna start with my derivatives over here. Now using the power rule on this s and then knowing the derivative of any constant is zero, my first derivative is just going to be 2s. Now my second derivative with respect to s is just going to be two. And then my third derivative with respect to s, well, that's going to be zero because the derivative of any constant is just zero. So now we've gotten this all the way down to zero, so I'll go ahead and stop right about there. Now let's move over to my integral column. Well, here we have the integral of e to the three s. What is that integral going to be? Well, if you integrate this, what you would want to do is a variable substitution on the three s right there. So if I'm integrating e to the three s with respect to s, well, what I'm gonna go ahead and do first is I'm going to take this three s here, I'm going to set it equal to u. So we'll say u equals three s. Now the derivative of u with respect to s is going to be three d s, just like so. So what I can do is say that this is now the expression e to the u d s, and what I can recognize is that d u is equal to three d s. So if I take this d s and I multiply it by three, that's going to be equivalent to the du, but then I need to divide it by three to offset this three that I randomly multiplied inside the integral. So doing that, I can take this three in the denominator, pull it out front of the integral right there, and then it's going to give me the integral of e to the u times three d s, which we know is just du. We know the integral of e to the u is just gonna come out to e to the u plus c, where I can go ahead and put this one third out front right here, and I can also take this u that I see as three s and replace the u right there. So this is how you can go ahead and deal with this integral right here. So as you can see, going through this process, we are able to figure out what our integral becomes. And so I can say that this e to the three s here is just going to become one third e to the three s. Now the cool thing is as you continue to take this integral, you're going to notice a pattern here. This three always just ends up getting divided on the outside of the e. If you go ahead and integrate this and then you keep going through that integral, so our next piece is going to be one ninth e to the three s, multiplying this three by the three down there. And then integrating again, we would get one over in this case, it would be three times nine, which is 27 e to the three s. So that's what happens when we work through these integrals and these derivatives, and now we can go ahead and go through with the problem. Now what's gonna happen with your derivatives column is the signs are going to alternate. So you're going to have a plus sign, a positive right here, then you'll have a negative, then you'll have a positive, and then you'll have a negative. So that's just gonna keep alternating. And then what you wanna do is start with what you have in your derivative column, and you're going to go diagonally down and to the right to the integral right there. So writing out this over here, we're going to have this times that. So that's going to be one third e to the three s multiplied by s squared plus four. So that's what we get first. Now next, we're going to move down here. I can see that we end up getting a negative sign over there, so that's going to be minus, and then we're gonna have two s times one over nine e to the three s. So So it's going to give us two over nine s e to the three s, multiplying those two. And then I'm going to go diagonally down into the right again. So we're going to have two times this piece, and I'm gonna go ahead and give myself some more space over here as we continue to write this out. So multiplying the two and the one over 27, that's going to give us and we'll have plus. So we have plus right there. That's gonna be two over 27. Then we'll have that e to the three s right there. And then what I need to do is move directly across right here. This isn't going directly across. Well, that's now going to be in my integral. So it's going to be minus and then the integral of zero times this whole thing. But that's just going to be zero anyway, so we don't even need to worry about the rest of this. So I just need to write plus the cost of integration c, and that right there is going to be the solution to my problem. So this is how you can use a table to solve these integration by parts cases. And notice how oftentimes this can be a lot quicker than just using the formula over and over again. That's why that it it's good to know all these various strategies we can use when it comes to dealing with integration by parts and how we can set things up, how we can get things more simple, because this often is gonna be a long process in any case. But knowing methods like this one with the table can allow us to do this in a more organized and quick fashion. So hope you found this video helpful. Let me know if you have any questions, and let's move on.

Evaluate the indefinite integral. ∫ ( s 2 + 4 ) e 3 s d s \int_{}^{}\left(s^2+4\right)e^{3s}ds

e 3 s 3 ( s 2 + 4 ) − 2 9 s e 3 s + 2 27 e 3 s + C \frac{e^{3s}}{3}\left(s^2+4\right)-\frac29se^{3s}+\frac{2}{27}e^{3s}+C

e 3 s 3 ( s 2 + 4 ) − 2 9 s e 3 s + C \frac{e^{3s}}{3}\left(s^2+4\right)-\frac29se^{3s}+C

( s 3 + 4 3 s ) e 3 s + C \left(s^3+\frac43s\right)e^{3s}+C

( s 3 3 + 4 3 s ) e 3 s + C \left(\frac{s^3}{3}+\frac43s\right)e^{3s}+C

12. Techniques of Integration

Integration by Parts

Calculus

12. Techniques of Integration

Integration by Parts

Struggling with Calculus?

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Multiple Choice

Evaluate the indefinite integral.
$\int_{}^{} (s^{2} + 4) e^{3 s} d s$

$\frac{e^{3 s}}{3} (s^{2} + 4) - \frac{2}{9} s e^{3 s} + \frac{2}{27} e^{3 s} + C$

$\frac{e^{3 s}}{3} (s^{2} + 4) - \frac{2}{9} s e^{3 s} + C$

$(s^{3} + \frac{4}{3} s) e^{3 s} + C$

$(\frac{s^{3}}{3} + \frac{4}{3} s) e^{3 s} + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a product of two functions, $ (s^2 + 4) $ and $ e^{3s} $. This suggests that we should use the integration by parts method. Recall the formula for integration by parts: $ \int u \cdot dv = u \cdot v - \int v \cdot du $.

Step 2: Choose $ u = s^2 + 4 $ (since it simplifies upon differentiation) and $ dv = e^{3s} ds $ (since it integrates easily). Compute $ du $ and $ v $: $ du = 2s \cdot ds $ and $ v = \frac{e^{3s}}{3} $ (integrating $ e^{3s} $ with respect to $ s $).

Step 3: Substitute into the integration by parts formula: $ \int (s^2 + 4)e^{3s} ds = (s^2 + 4) \cdot \frac{e^{3s}}{3} - \int \frac{e^{3s}}{3} \cdot 2s \cdot ds $. Simplify the first term and focus on the remaining integral.

Step 4: For the remaining integral $ \int \frac{2s e^{3s}}{3} ds $, apply integration by parts again. This time, let $ u = s $ and $ dv = \frac{2e^{3s}}{3} ds $. Compute $ du = ds $ and $ v = \frac{2e^{3s}}{9} $. Substitute into the formula and simplify.

Step 5: Combine all terms from the integration by parts steps, ensuring to simplify coefficients and constants. Don't forget to add the constant of integration $ C $ at the end. The final expression will involve terms like $ \frac{e^{3s}}{3}(s^2 + 4) $, $ -\frac{2}{9}se^{3s} $, and $ \frac{2}{27}e^{3s} $.

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