Using accumulation functions 2015 AP Calculus AB BC 1a

0:00- [Instructor] The rate at which rainwater flows into a drain pipe is modeled by the function R, where R of T is equal to 20 sine of T squared over 35 cubic feet per hour. T is measured in hours and zero is less than or equal to T, which is less than or equal to eight, so T is gonna go between zero and eight.

0:21The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by D of T. It's equal to negative 0.04T to the third power plus 0.4T squared plus 0.96T cubic feet per hour, for the same interval right over here. There are 30 cubic feet of water in the pipe at time T equals zero.

0:47All right. Part A. How many cubic feet of rainwater flow into the pipe during the eight hour time interval zero is less than or equal to T is less than or equal to eight. All right, so we know the rate that things flow into the rainwater pipe. In fact, we could, let me draw a little rainwater pipe here just so we can visualize what's going on.

1:12So if that is the pipe right over there, things are flowing in at a rate of R of T and things are flowing out at a rate of D of T. And they even tell us that there's 30 cubic feet of water right in the beginning, but these are the rates of entering and the rates of exiting.

1:35So how much water, they're asking how many cubic feet of water flow into, so enter into the pipe, during the eight hour time interval. So if you have your rate, this is the rate at which things are flowing into it. They give it in cubic feet per hour. If you multiply it times some change in time, even infinitesimally small change in time, so DT, this is the amount that flows in over that very small change in time.

2:03And so what we wanna do is we wanna sum up these amounts over very small changes in time to go from time is equal to zero all the way to time is equal to eight. So this expression right over here, this is going to give us how many cubic feet of water flow into the pipe.

2:23Once again, what am I doing? R of T times D of T, this is how much flows, what volume flows in over a very small interval of DT. And then we're gonna sum it up from T equals zero to T equals eight. That's the power of the definite integral. And so this is going to be equal to the integral from zero to eight of 20 sine of T squared over 35 DT.

2:53And lucky for us, we can use calculators in this section of the AP exam. So let's bring out a graphing calculator where we can evaluate definite integrals. And so let's see, we wanna do definite integral so that I can click math right over here, move down. So this function, FN integral, this is a integral of our function or function integral right over here.

3:16So let me press enter. And the way that you do it is you first define the function, then you press put a comma, then you say what variable is the variable that you're integrating with respect to, and then you put the bounds of integration. So I'm gonna write 20 sine of, and just 'cause it's easier for me to input X than T, I'm gonna use X, but if you just use this as sine of X squared over 35 DX, you're gonna get the same value.

3:43So you're going to get X squared divided by 35, divided by 35, close that parenthesis. So that is my function there. Actually, I don't know if it's going to understand, I don't know if it's going to understand, let me put the times, so second insert times just to make sure it understands that.

4:08Okay, so that's my function. And then let me throw a comma here, make it clear that I'm integrating with respect to X. I could have put a T here and integrated with respect to T, we'd get the same value, comma, my lower bound is zero and my upper bound is eight and close the parentheses and then let the calculator munch on it a little bit and we get 76.570.

4:35So this is approximately 76.570.