Find the integral.∫se3sds\int_{}^{}se^{3s}ds∫se3sds

s 3 e 3 s − 1 9 e 3 s + C \frac{s}{3}e^{3s}-\frac19e^{3s}+C 3 s e 3 s − 9 1 e 3 s + C

Find the integral. ∫ s e 3 s d s \int_{}^{}se^{3s}ds ∫ s e 3 s d s

give this problem a try. We're asked to find the integral of s times e to the three s with respect to s. Now how do I go about solving this problem? Well, something that I notice is that we have a product of two functions right here, and we're integrating them. Now you could try doing a variable substitution, like you could try a variable substitution here on the three s, but no matter where you tried this, it turns out you're actually not going to get things to be more simple, because if you did a variable substitution here, when you take the derivative, it would actually clear the s, so therefore, you would not be able to simplify things with the s out in front. So that means we're going to need to use integration by parts. Now we've learned the strategy for this general process, and so far, what we've been doing is applying this formula. The integral of u d v is equal to u times v minus the integral of v d u. Now, of course, our step here is figuring out what exactly our u and our d v are. Well-to-do that I can go over to my integral right here. Now looking at this the u needs to be something that becomes more simple when you take its derivative. Now looking at this I can see the derivative of s is just gonna be one, so I'm going to say u equal to s, and that means the derivative of u is gonna be the derivative of s, which is just one ds. Now next I need to find dv. Dv needs to be something that can easily be integrated. Now notice that we have this e to the s right here, or this e to the three s, I should say. This e to the three s, I know that can be easily integrated because e to the three s with respect to s, well, integrating that is just going to give me one over three times e to the three s. Now if you're wondering how to do this integral right here, what you could do is do a u substitution or a variable substitution up here, and then when you take its derivative, it would give you this one third that you could get out in front, and then it would give you e to the three s that keeps that right there. But that would be how you could go ahead and find your v when you go ahead and integrate this function. So now that we've found u and du and d v and v, we can go ahead and apply this formula in full. So first off we're going to have u times v out in front, which is going to be s times one third e to the three s, you'll have minus the integral that's going to be a v du. Now v in this case is one third e to the three s, and du well that's just going to be this ds right here. So now we've gone ahead and applied this. Now what I can do from here is try to simplify things. So this right here, s times one third e to the three s, I'm just gonna write that as s over three e to the three s minus, and then we're going to have I'm gonna pull this one third outside here, so we're gonna have one third times the integral of e to the three s with respect to s. Now at this point, the integral of e to the three s, that's something I know how to do. Remember, that's just gonna be one third e to the three s. We've done that integral already. So if I wanna go ahead and simplify this, we're gonna have s over three e to the three s minus one third, and then it's going to be the integral of e to the three s which is just one third e to the three s plus the constant of integration c. Now at this point I can multiply the one third times one third to give me one ninth, so we're going to have s over three times e to the three s minus, and then let's go write the s a little bit or minus, then we'll have one ninth e to the three s plus the constant integration c, and that right there is the solution to the problem. So this is how you can use integration by parts to solve this problem. Hope you found this video helpful, and let's move on and get some more practice.

Find the integral. ∫ s e 3 s d s \int_{}^{}se^{3s}ds ∫ s e 3 s d s

s 3 e 3 s − 1 9 e 3 s + C \frac{s}{3}e^{3s}-\frac19e^{3s}+C 3 s e 3 s − 9 1 e 3 s + C

s 9 e 3 s − 1 3 e 3 s + C \frac{s}{9}e^{3s}-\frac13e^{3s}+C

3 s e s − 1 9 e 3 s + C 3se^{s}-\frac19e^{3s}+C 3 s e s − 9 1 e 3 s + C

3 s e 3 s − 9 e 3 s + C 3se^{3s}-9e^{3s}+C 3 s e 3 s − 9 e 3 s + C

11. Techniques of Integration

Integration by Parts

Business Calculus

11. Techniques of Integration

Integration by Parts

Struggling with Business Calculus?

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Multiple Choice

Find the integral.
$\int_{}^{}se^{3s}ds$

$\frac{s}{9} e^{3 s} - \frac{1}{3} e^{3 s} + C$

$3se^{s}-\frac19e^{3s}+C$

$\frac{s}{3}e^{3s}-\frac19e^{3s}+C$

$3se^{3s}-9e^{3s}+C$

Verified step by step guidance

Step 1: Recognize that the integral ∫se^{3s}ds involves a product of two functions, s and e^{3s}. This suggests using integration by parts, which is based on the formula ∫u dv = uv - ∫v du.

Step 2: Choose u = s (the polynomial term) and dv = e^{3s}ds (the exponential term). Then, compute du = ds and v = ∫e^{3s}ds. To find v, integrate e^{3s} with respect to s, which gives v = (1/3)e^{3s}.

Step 3: Apply the integration by parts formula. Substitute u, v, du, and dv into ∫u dv = uv - ∫v du. This results in ∫se^{3s}ds = s * (1/3)e^{3s} - ∫(1/3)e^{3s}ds.

Step 4: Simplify the remaining integral ∫(1/3)e^{3s}ds. Factor out the constant (1/3) and integrate e^{3s}, which gives (1/9)e^{3s}. Substitute this back into the equation.

Step 5: Combine all terms to express the solution as (s/3)e^{3s} - (1/9)e^{3s} + C, where C is the constant of integration.

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