Negative definite integrals | Integration and accumulation of change | AP Calculus AB | Khan Academy

We've already thought about what a definite integral means. If I'm taking the definite integral from ( a ) to ( b ) of ( f(x) , dx ), I can just view that as the area below my function ( f ).

So, if this is my y-axis, this is my x-axis, and ( y ) is equal to ( f(x) ). Something like that: ( y ) is equal to ( f(x) ). And if this is ( a ), and if this is ( b ), I could just view this expression as being equal to this area.

But what if my function was not above the x-axis? What if it was below the x-axis? So, these are going to be equivalent. Let's say, let me just draw that scenario. So let me draw a scenario where it's my x-axis that is my y-axis. And let's say I have, let's say I have a function that looks like that.

So that is ( y = g(x) ), and let's say that this right over here is ( a ), and this right over here is ( b ). And let's say that this area right over here is equal to 5. Well, if I were to ask you, what is the definite integral from ( a ) to ( b ) of ( g(x) , dx )? What do you think it is going to be?

Well, you might be tempted to say, "Hey, well it's just the area again between my curve and the x-axis." You might be tempted to say, "Hey, this is just going to be equal to 5." But you have to be very careful because if you're looking at the area above your curve and below your x-axis versus below your curve and above the x-axis, this definite integral is actually going to be the negative of the area.

Now, we'll see later on why this will work out nicely with a whole set of integration properties. But if you want to get some intuition for it, let's just think about velocity versus time graphs.

So if I... in my horizontal axis, that is time. My vertical axis, this is velocity, and velocity is going to be measured in meters per second. Time is going to be measured in seconds. Time is measured in seconds, and actually I'm going to do two scenarios here.

So let's say that I have a first velocity-time graph. Let's just call it ( v_1(t) ) which is equal to 3, and it would be 3 meters per second. So one, two, three. So it would look like that: that is ( v_1(t) ). And if I were to look at the definite integral going from time equals 1 to time equals 5 of ( v_1(t) , dt ), what would this be equal to?

Well, here my function is above my ( t )-axis, so I'll just go from 1 to 5, which will be around there. I could just think about the area here, and this area is pretty easy to calculate. It's going to be 3 meters per second times 4 seconds; that's my change in time.

So this is going to be 12 meters, and so this is going to be equal to 12. One way to conceptualize this is this gives us our change in position. If my velocity is 3 meters per second and since it's positive, you can conceptualize that as it's going to the right at 3 meters per second.

What is my change in position? Well, I would have gone 12 meters to the right, and you don't need calculus to figure that out. ( 3 , \text{m/s} \times 4 , \text{s} ) would be 12 meters. But what if it were the other way around? What if I had another velocity function?

Let's call that ( v_2(t) ) that is equal to negative 2 meters per second, and it's just a constant negative 2 meters per second. So this is ( v_2(t) ) right over here. What would or what should the definite integral from 1 to 5 of ( v_2(t) , dt ) be equal to?

Well, it should be equal to my change in position. But if my velocity is negative, that means I'm moving to the left. That means my change in position should be to the left as opposed to to the right.

So we can just look at this area right over here. When if you just look at it as the rectangle, it is going to be ( 2 \times 4 ), which is equal to 8. But you have to be very careful since it is below my horizontal axis and above my function. This is going to be negative.

And this should make a lot of sense. If I'm going 2 meters per second to the left for four seconds, or another way to think about it, if I'm going negative two meters per second for four seconds, then my change in position is going to be negative eight meters. I would have moved eight meters to the left if we say the convention is negative means to the left.

So the big takeaway is if it's below your function and above the horizontal axis, the definite integral, and if your ( a ) is less than ( b ), then your definite integral is going to be positive. If your ( a ) is less than ( b ), but your function over that interval is below the horizontal axis, then your definite integral is going to be negative.

And in the future, we'll also look at definite integrals that are a mix of both, but that's a little bit more complicated.