# Explore of time. Let’s say your velocity is

Explore how driving backwards takes you where you’ve already been as we define definite integrals. This lesson will also teach you the relationship between definite integrals and Riemann sums.

Then, discover how an integral changes when it is above and below the x-axis.

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## Definite Integrals

Let’s say you want to find the area between some line and the x-axis. You know that you could use a Riemann sum approach. This approach divides the region up into a bunch of different slices, and you estimate the area of each slice. As the width of each slice gets thinner, your estimate of the total area gets better.

Eventually, you’ll get the total area exactly. We write this as the limit as delta x goes to zero of the sum of all of the slices from k=1 to n of f(x sub k) * delta x sub k, or the height times the width of each slice.This limit equals the definite integral from a to b of f(x)dx. This is the integral, but what does integral mean? We’ve said that it can mean the area under the curve, but does it always mean that?

## Example #1

In this case, the integral of velocity as a function of time gives you the area under the curve, which is your distance traveled.

We can make this a little more specific and say that if your velocity is given as a function, f(t), and you’re traveling from time a to time b, then the distance that you’ve traveled equals the integral from lower limit a to upper limit b of f(t)dt. In this case, you’re integrating the function f(t) with respect to t.

## Example #3

But what happens in a case like this? Here, your velocity is 30 mph for the first half hour – maybe you’re driving away from home – and 30 minutes in, your velocity becomes -30 mph. You stop the car a half hour into your drive, put it in reverse and drive backwards at 30 mph for a half hour. In this case, if you go forward at 30 mph for a half hour and then you go backward at 30 mph for a half hour, you’re going to end up in the same place.

So here if I integrate from a lower limit of t=0 to an upper limit of t=1 my function f(t)dt, I get zero. If I were just looking at the area, it wouldn’t be zero, it would be a positive number. What happened?This is a very important fact about integrals. Before we said that a Riemann sum is just an area, but an integral can be positive or negative. If you’re above the x-axis, it’s the area, but if you’re below the x-axis, it’s the negative area.

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