Definite Integrals As Net Change - Problem 1
Let's do a Net change theorem problem. First of all, recall that net change theorem is basically another version of the fundamental theorem of Calculus. The integral from a to b of F'(x) is F(b) minus F(A). We can use integral to calculate a net change of some function F. Well, let's see how it works out in this problem.
Heating oil is being pumped into a tank at a rate in gallons per minute given by A'(t) equals 40 minus 8 root t; for t between 0, and 25 where t is the time in minutes since the pumping began. If the tank started off with 100 gallons of heating oil, how much will contain 9 minutes after the pumping began.
So remember its starting off with 100 gallons. All I know about the pumping is the rate at which the oil is being pumped into the tank. I don't know actual amounts. Well, first of all, let's write this down; A'(t) 40 minus 8 root t. This is for t between 0, and 25. This is the rate that oil's being pumped into the tank. So rate of change of the amount of oil in the tank.
Now let's just remember that this had units of gallons per minute. That's important because if we want to figure out what A(t) is going to be like. A(t) is my function for the amount of oil in the tank. This is going to be in gallons. A'(t) is gallons per minutes. Minutes are the units of time here. A(t) is in gallons.
So I also know A(0) is 100. We start of with a 100 gallons. I need to find the amount after 9 minutes. I need to find A(9). So in this kind of situation, you could use the net change theorem.
I know everything I need to know about the rate of change of the amount. I know the initial amount, and I need to find this final amount. This net change A(9) minus A(0) according to the net change theorem equals the integral from 0 to 9 of A'(t)dt. So I'm going to use this integral to calculate A(9). A(9) minus 100. A(0) is 100. A' is 40 minus 8. I'm going to write t to the 1/2 for integration purposes.
I need an antiderivative for this. So let me take this whole problem up here. I'm going to write A(9) minus 100 equals. The antiderivative, going back here, for 40 minus 80 to the 1/2 is 40t minus, 8 times an antiderivative of t to the 1/2, is t to the 3/2 divided by 3/2. So 2/3t to the 3/2. 8 times that is 16/3t to the 3/2. So 40t minus 16/3 t to the 3/2. I'm going to evaluate this from 0 to 9.
First at 9. When I plug 9 in I get 40 times 9 which is 360 minus 16/3. 9 to the 3/2. First I take the square root of 9 I get 3, 3³ is 27. So it's 16/3 times 27 minus. When I plug 0 into these two guys, I get 0.
So I have 360 minus 16/3 of 27, that's 16 times 9 which is 144. 360 minus 144, that's 216 gallons. But I'm not done yet. Remember I'm trying to find A(9), and I had to subtract 100. So this is A(9) minus 100. So A(9) is 316 gallons. So what did I just learn? Over that 9 minute period, 216 gallons of oil have been poured into the tank. It started off with 100 gallons, so it now has 316 gallons of oil.