# 6 – Nd113 C6 07 L The Integral Area Under A Curve H1 V2

We’re going to come back to the elevator example in a little bit. First, I want to remind you of something you saw a few of courses back. Earlier in this category, when you were learning about matrices in motion models, you may remember seeing a graph that looks something like this. And this was a graph of velocity in meters per second versus time in seconds. And back then, Suzanne told you that if you want to calculate the displacement or distance moved between T1 and T2, you could do so by finding the area under this curve. And in this case, if T1 were two and T2 were four, well that would mean that the width of this rectangle is just four minus two which is two and the height of the rectangle, that’s just one 100m/s. I can use this equation, area = width * height to calculate that area. And in this case, that would be two seconds, times 100m/s and with it written out this way, you can see something interesting. The seconds actually cancel and all I’m left with is this, an area of 200 meters. And now you can see why I’ve been so focused on the unit’s associate with motion, because if you look at the units for the axes on this graph, you can see that the vertical axis has units of meters per second, and the horizontal axis has units of seconds. And when you multiply these units together, you’re just left with meters. Whenever you calculate an integral, you’ll be doing some sort of multiplication just like this and you’ll be looking at the units on the axis of your graph to get a hint for what the integral or area under the curve actually represents. Now I’d like you to think about a few questions involving graphs of motion and areas under the curve.