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One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. (The photo shows a screw-worm fly, the first pest effectively eliminated from a region by this method.)

Let $ P $ represent the number of female insects in a population and $ S $ the number of sterile males introduced each generation. Let $ r $ be the per capital rate of production of females by females, provided their chosen mate is not sterile. Then the female population is related to time $ t $ by$$ t = \int \frac{P + S}{P [(r - 1) P - S]}\ dP $$Suppose an insect population with 10,000 females grows at a rate of $ r = 1.1 $ and 900 sterile males are added initially. Evaluate the integral to give an equation relating the female population to time. (Note that the resulting equation can't be solved explicitly for $ P $.)

$$\ln \frac{10,000}{P}+11 \ln \frac{P-9000}{1000}$$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Integration Techniques

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Let's go ahead and evaluate the answer rule that's given So here, Looking at this into Grand, we should do a partial fraction decomposition. So let's go to the side and write this. So, using what the author calls Case one, here's a partial freshen. Now it's good and multiply both sides of this equation by the denominator on the left. So we have P plus this. Then we have a are minus one p minuses. Be happy. Got a fat therapy and there are constants. Verma's es with a minus outside. Here we see the coefficient in front of the piano left is one. It's on the right hand side. That must be one. We also see the coefficient firm. V S is the one. So that means that negative a was one. So we end up with a equals one and plugging that a value and over here and solving for B, we can be equals. R. Let's plug in These values for ambience are partial fraction decomposition and then we'LL integrate this term instead of the original. So going head and plugging those in and then that's going to evaluate these Now this second Integral is bothering you. You can go ahead and go use up here. And then that should lead you to this in a roll over here. So PMS or Constance? Well, this is a constant here. So going on and now plugging in our values for arness, recall ours. One point one s is nine hundred. These were given so c equals negative lmp. He's positive. So if you could drop the absolute value plug in our than our mind, this one that's a point one down there and then natural log. Absolute value are minus one is point one and then times p and then minus nine hundred. Absolute value plus e. Now, also, we are given that when t equals zero, he is ten thousand. So we can use this information to find see so plugging in t zero into the left hand side. We have a zero here and then on the right. Negative Ellen, ten thousand. And then we have go ahead and simplify this fraction. You just get eleven natural log and then one thousand minus that hundred after groups at nine hundred. After plugging in Pete plus he and then go ahead and isolate and solve for C That's our see value. So finally, for tea, we have negative natural log thi and then eleven natural log point one P minus nine hundred. It's a point one out there, so one over ten in the front and find the pea. Only Ellen of ten thousand minus eleven. Natural log of one hundred. You know, simplify this Helen of ten thousand overpay. So I'm just using combining these two and using a lot of properties. And then I could pull out in eleven and use a lot of properties again here. So on top will have quite one times p minus nine hundred over one hundred. And then we could clean this second expression of a little bit. So here, luscious God, and write thie inside the absolute values. So we have eleven natural log. So here, let's just go ahead and multiply top and bottom by ten. And there's a final answer

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