💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # The rates at which rain fell, in inches per hour, in two different locations $t$ hours after the start of a storm are given by $f(t) = 0.73t^3 - 2t^2 + t + 0.6$ and $g(t) = 0.17t^2 - 0.5t + 1.1$. Compute the area between the graphs for $0 \le t \le 2$ and interpret your result in this context.

## 0.867in

#### Topics

Applications of Integration

### Discussion

You must be signed in to discuss.
##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University  ##### Kristen K.

University of Michigan - Ann Arbor ##### Michael J.

Idaho State University

Lectures

Join Bootcamp

### Video Transcript

the rates at which rain fell in inches per hour in two different locations. T hours after the start of a storm are given by the falling functions here, FFT, which is going to be in Green, and GFT, which is going to be in blue. What we want to do is compute the area between these two graphs for the time interval, when tea is between zero and two hours and we want to interpret our results in context. So the first thing to do since we want to compute area between two functions is since we're going to do an integral, we have to decide which function is on top and which is on bottom because we have to know which one to subtract. So let's plug in some points and see what our results are. So let's start with the boundary point. Let's suppose that T is equal to zero so plugging, and for F, we have F zero Well, that gives us 0.6. Now let's compare that Thio G of zero g of zero gives us 1.1, so let's begin graphing this So F is at 0.6 when tea is zero. Now, the way that I've constructed this coordinate system is that my Y axis is the rate and my X axis is the time. And so we're about right there. All right, now, let's check. Oh, let's suppose t is one. We're at one hour. So for our f function f of one, that gives us 0.33 and g of one results in 0.67 So let's pluck those yet again. So f is our green and where it 0.33 and G is our blue and we're at 0.67 so far it looks like blue is above. Let's just check one more point. So how about the end point and boundaries? So too f of two. That gives us 0.44 n g of two gives us one point. Uh, excuse me. 0.78 Okay, So were to wear it 0.44 for our dream. And we're at 0.78 for our blue. So roughly this kind of sketch And so we can now see pretty clearly which one is on top in which one is on bottom. So our GFT is going to be our first function and we will be subtracting FFT from GFT. Our boundary is our T values which were given that tea is in between zero and two. So let's construct the integral. So we have are integral from 0 to 2 of f g of T minus fft g of tea minus f of tea duty. So that is equal to the integral from 0 to 2 of 0.17 t squared minus 0.5 t plus 1.1 that is our GFT minus 0.73 t cubed minus to t squared plus T plus 0.6 and all of that DT with respect ity. So let's compute the integral. Um, well, first, let's combine like terms with inner integral. So distributing that negative through this gives us a negative 0.7 to 3 t cute. It gives us a positive T squared, a negative T and a negative zero point 0.6. Okay, so then combining these we have the integral from 0 to 2 off negative 0.73 t cute writing the powers in descending order plus 2.17 t squared minus 1.5 T plus 0.5 Duty. All right. Now evaluating the integral. From there, we get negative 0.73 over four T to the fourth. Plus to over 72.17 over three t cute, minus 1.5 over too. T squared, plus 0.5 I see evaluated from 0 to 2. Now we plug in to into each term, find that value and then subtract the resulting value that we get after we plug in our lower bound zero. So for plugging into, we would have negative 0.73 over four time, 16 plus two points, 17 over three times eight minus 1.5 over too. Times four plus 0.5 times two. And then when we subtract the resulting value after playing in our lower bound zero Well, all of that just tends to zero. So we're subtracting zero. Plug this into your calculator. Of course, you could do a little bit of simplifying. Four cancels with 16 to give us four to cancels with four to give us to et cetera, and you get zero point 8666 weekend around that to 0.867 Now recall we're finding the area between these two curves the blue curve and the green. Kurt, we're finding the area from zero 22 So what this is, is we're taking inches per hour, times, hours. Think of this. Instead of being like a to Bess, it is. Think of it as a square. Right you went, you would take if it were like a square, you would take the bottom and you'd multiply the side just like you'd find the area of a square, which is length, times, width. And that gives you your your unit's times, your units. So it's the same idea that we're doing here. Except it's not a square, but it's the same idea. It'll be inches per hour, times, hours, the hours cancel, and we're just left with inches as our result. So there's a red in here is in the units of inches, so this is a net, um, amount or a net distance. However you want to say it, so our units here would be in inches. So since we were talking about the rate of rainfall, this is the net change in amount of rainfall, so net change in amount of rainfall. Zero point 86 867 inches Missouri State University

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators ##### Catherine R.

Missouri State University  ##### Kristen K.

University of Michigan - Ann Arbor ##### Michael J.

Idaho State University

Lectures

Join Bootcamp