The rate of growth of a particular bacteria is given by b ′ ( t ) = 4 t 2 e .5 t b^{\prime}\left(t\right)=4t^2e^{.5t} where t t is time in days. What is the total growth of the population of bacteria during the first 5 days?

The rate of growth of a particular bacteria is given by b prime of t equals four t squared times e to the 0.5 t, where t is the time in days. What is the total growth of the population of bacteria during the first five days? Okay. So this is an interesting problem. And to solve it, what we need to do is take a look at what we're given in this problem and figure out where we're trying to go. Now what I notice is that we are given a rate of growth in this problem. That's b prime of t is equal to four t squared times e to the 0.5 t. Now a rate is basically just a derivative, because that's like a rate of change. And if I'm given a rate, and I'm looking for the total growth, so I'm given a rate of growth, and I'm looking for total growth, I think it makes sense that all I really need to do is integrate my function for the rate of change, because an integral is just the opposite of a derivative. So what I'm going to do is go ahead and integrate this function that we have, four t squared times e to the 0.5 t, and then from here, I'm going to bound this for the time interval that we're looking for. In this case, it's the first five days, so I'm gonna bound this from zero to five. So now I have this integral set up. Now at this point, what we need to do is figure out how we're gonna go about solving this integral. Now looking at this, I can see that I have a product of two functions, four t squared and e to the 0.5 t. Now it may be tempting to try a variable substitution at this point, and you're always welcome to try this when it comes to these problems. But if you do a variable substitution on the 0.5 t, you're gonna have to take the derivative of that. And the derivative of 0.5 t with respect to t is just 0.5. Now the reason that's not gonna help us much is because we have a four t squared out in front, so taking that derivative is not going to get anything to cancel or simplify things with that four t squared. Now alternatively, you could try u substitution right here on the four t squared, but when you take that derivative, it's not going to help you cancel anything that's in this argument of the function of e because we have e to an exponent of 0.5 t, which isn't going anywhere. So because the variable substitution isn't gonna help us here, we're gonna need to use integration by parts. Now we should be familiar with this formula and the general process for integration by parts, and our first step is typically gonna be figuring out what this u is. U needs to be more simple when after we take the derivative. Now looking at this function of e here, I know that if I take the derivative of e to the 0.5 t, I'm just gonna get another argument with e in it. That's not really gonna help me. So instead, what I'm gonna do is take a look at this four t squared. If I take the derivative of 4t squared, that's going to give me 8t. That does seem more simple overall, so I'm going to say that my u is 4t squared, and I know the derivative of u is just going to be the derivative of 4t squared, which is 8t, with respect to t. So now we've found our u and our du. Now next I need to find my dv. Dv needs to be something that's easily integrated, and I know that these functions of e are pretty straightforward to integrate. So I'm going to say that e to the 0.5 t is going to be my dv. And to find v, well, I just need to go ahead and integrate 0.5 t. How do I go about doing that? Well, the integral of e to some function of t, well, what you need to do is look at whatever is in front of this t, or in whatever this variable might be. Since I can see it's 0.5, that's going to be divided on the outside. So I have one over 0.5 times the same function e to the 0.5 t. And one over 0.5, well, that's just the same thing as two. So that's two e to the 0.5 t, and that's what our v is going to be. Now if you weren't sure how I went ahead and went through this process or figured this out, what you could do is the long way of doing this integral. So if you wanted to integrate e to the 0.5 t with respect to t, you could first do a variable substitution on this 0.5 t. Then doing that, you take the derivative, which would give you 0.5, and then that means you'd have to multiply this by 0.5, then divide by 0.5, bringing this out to the front, thus giving you this result right here. So you could go through the entire variable substitution to see how that all works, but I wanted to go ahead and give you the quick answer here so this is our v and our dv, this is our u and our du, and now we have the variables that we need so we can go ahead and use this equation. So what we're first going to have is u times v, so that's going to be four t squared times two e to the 0.5 t, then we're going to have minus and that's going to be the integral of v du. So we're gonna have v times du, which means we're going to have well, I'll actually multiply these two right here. So we're gonna have two times e to the 0.5 t times this function right here, eight t dt. And then keep in mind, this is the definite integral we're dealing with, so this is gonna go from zero to five, and this piece is also gonna go from zero to five. Both parts you need to bound with what you have on your integral. So now we have this set up. Now before applying the bounds right here, I'm gonna go ahead and see if I can solve this part of the integral first. So I'm just gonna go ahead and keep these bounds on the 14, and what I will do is combine these constants actually. So we have the two and the four right there, the four and the two, so I'll combine those to give me eight. So we'll get eight t squared, and that's gonna be e to the 0.5 t. All of this is bounded from zero to five, and then we'll have minus, and then I need to deal with this integral. Now this integral is going to be the same thing as two times eight, which is 16, and that's gonna be times the integral from zero to five of t times e to the 0.5 t dt. Just arranging this all to look a little nicer. Now notice when we do this, we get another product of two functions that shows up, And this is another product of two functions where you actually couldn't do a variable substitution. The variable substitution here would not help us with what's in the exponent of e. Doing the variable substitution here would not get that t out in the front to cancel. So we're gonna have to use integration by parts a second time. So what that means is that I need to go through this process again. So to do this, well, this whole piece is gonna say that today, eight t squared times e to the 0.5 t bounded from zero to five. Then we have minus 16 times our next portion for integration by parts, which is gonna be everything that we have right here. So I need to find another u. Well, to do this, u is going to be t. Because I know that when I went ahead and set u to the function of t squared, four t squared, that caused things to be a lot more simple here. So I'm gonna go ahead and say that that t is equal to my u once again, because I know that that's just gonna get more simple. So if I go ahead and take this derivative, the derivative of t is gonna be just one dt. And notice that does get me to a more simple result right there. So we found our next u and our next du, and then next I'm gonna find my dv. Well, dv is once again gonna be this argument or this function of e to the 0.5 t, and we know that this integral just comes out to two e to the 0.5 t. So now we've found our u, du, our dv, and v, so you can go ahead and use this formula once again. So going right here to this formula applied second time we're going to have u times v which is going to be t times two e to the 0.5 t, all that's going to be bounded from zero to five, then we have minus the integral from zero to five of v du, which is gonna be two e to the 0.5 t dt. So this is what we get. Now again, I'm still gonna wait to apply these bounds before I've gone ahead and simplified this as far as I can get. So we'll keep this eight t squared e to the 0.5 t out in front. We'll have minus 16 times, and then we're going to have in this case, I'll simplify things, so we'll get two t e to the 0.5 t. Just writing this in a more arranged way so it's a little easier to read, and then I'll go ahead and factor this to or pull this to I should say outside of the integral right here, so I have zero to five and that's e to the 0.5 t integrated with respect to t. Now at this point I just need to go ahead and deal with this part of the integral, but I know how to integrate e to the 0.5 t. We've discussed that that just comes out to two e to the 0.5 t. So now we've gotten this to a point where I actually can integrate all of this. So this is going to be eight t squared e to the 0.5 t minus 16 times two t e to the 0.5 t bounded from zero to five minus, and then it's going to be two times, and then it's going to be two e to the 0.5 t since we know that's the result it comes out to, and all of this is going to be bounded from zero to five as well. So if I want to rewrite this all out, well, what I'm going to write this as is eight t squared e to the 0.5 t minus 16 times, and in this case we're going to have two t e to the 0.5 t, and it's going to be minus two times two which is four e to the 0.5 t. Just went ahead and combined those twos right there, and then all of this, all of this result that we have right here is all gonna be bounded from zero to five. Now I can go ahead and make this even more simple by taking this negative 16 and I can multiply it inside these brackets here. So we're going to get eight t squared e to the 0.5 t, that's going to be minus 16 times two, which turns out to be 32. We're gonna have 32 t e to the 0.5 t, and keep in mind this is going to be negative since we have that minus sign out in front. Then we have negative 16 times this negative four out in front here. Well, that's going to give me plus 64 e to the 0.5 t, and then all of this right here is going to be bounded from zero to five. So at this point, what I can go ahead and do is plug in my bounds. So first off, I'm gonna plug in my high bound of five. Well, doing this is gonna give me eight times five squared times e to the 0.5 times five minus 32 times five. That's gonna be times e to the 0.5 times five, plus 64 e to the 0.5 times five. Plugging five in for every place ICT, they were going to have minus, and then I'll plug zero in, my lower bound right here. So plugging zero in, well, that's going to give me, in this case, we'll have eight times zero. Well, this is just gonna set the whole thing equal to zero right here, and I'll have 32 times zero. Well, once again, that's gonna set this whole piece equal to zero. So the only place I really need to plug zero in is 64 e to the 0.5 times zero, and anything to the zero power is just gonna be equal to one, so we're really just gonna end up with this 64 right out here. So we're going to have all of this right here, all of this portion, we're going to go ahead and subtract off 64 from that result. So we can even remove these brackets here, and then you could just do this on a calculator and plug everything in directly across. So if you do that, you should get that your result approximately comes out to 1,203. So this right here would be the solution to the problem. So as you can see, this is the way that we can apply integration by parts to these word problems that we're dealing with. In this case, what we were looking for was the total growth of population during the bacteria for the first five days. And so we can say that about 1,203 of these bacteria was the growth rate based on the rate of growth that we were given in the first place. So that's how much it grew in the first five days, and this is how we could find the solution to our problem. So, hopefully, you found this video helpful. Let me know if you have any questions, and let's move on.

11. Techniques of Integration

Integration by Parts

Business Calculus

11. Techniques of Integration

Integration by Parts

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Multiple Choice

The rate of growth of a particular bacteria is given by $b^{'} (t) = 4 t^{2} e^{.5 t}$ where $t$ is time in days. What is the total growth of the population of bacteria during the first 5 days?

3,216

$1 comma 656$

$1 comma 203$

$2 comma 826$

Verified step by step guidance

Step 1: Understand the problem. The rate of growth of the bacteria is given by the derivative function b′(t) = 4t^2e^(0.5t). To find the total growth of the bacteria population during the first 5 days, we need to compute the definite integral of b′(t) from t = 0 to t = 5.

Step 2: Set up the definite integral. The total growth is represented by the integral: ∫[0 to 5] 4t^2e^(0.5t) dt.

Step 3: Use integration by parts if necessary. Since the integrand involves a product of a polynomial (4t^2) and an exponential function (e^(0.5t)), consider using integration by parts or substitution to simplify the integral.

Step 4: Evaluate the definite integral. After finding the antiderivative of the function, substitute the upper limit (t = 5) and the lower limit (t = 0) into the antiderivative and subtract the results to compute the total growth.

Step 5: Interpret the result. The value obtained from the definite integral represents the total growth of the bacteria population during the first 5 days.