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### Course: Class 11 Physics (India)>Unit 5

Lesson 3: Definite integral as area

# Worked examples: Definite integral properties 2

Sal evaluates definite integrals of functions given their graphs. He does so using various properties of integrals.

## Want to join the conversation?

• This is weird without any function equations, almost too easy. Is this just to get us used to the notation?
• Yes, there is a huge missunderstanding for integrals UNDER the x-axis, and I believe many teachers want to eliminate this by explaining it in the beginning of the topic.
• For anyone struggling with the notion of a "negative area", just think of this:

Imagine the graph in the video above is representing the velocity of a vehicle over a certain time. (velocity = y axis, time = x axis). Let's also consider only the right half of the graph (ie. only positive values of X).

By looking at the graph, you can see that at x = 0 (ie. when time is equal to 0), our velocity is still peaking, but soon it starts going down and finally reaches zero when time is equal to 3. Then we get a "negative" velocity, which essentially means our vehicle is moving in the other direction (it's "coming back", so to say). It's velocity keeps increasing in the opposite direction until roughly time = 5, then it starts decreasing again until it finally reaches zero again at time = 7.

The area under the function is supposed to give us the distance traveled. On the interval of time from 0 to 3 we crossed a positive distance, but then on the interval 3 to 7 we crossed a "negative" distance, ie. we came back around a certain distance. So to get the total distance traveled (ie. the "total area under the curve" as we call it), you have to treat the area under the X axis as if it were negative so that the distance covered during this period of time can be subtracted from the distance covered over the first (ie. where x is between 0 and 3) period of time.

So again, if we were to treat this as the graph of the velocity of a vehicle over the course of 7 hourse (x from 0 to 7), what the graph is essentially telling us is:

"From hour (0) to hour (3) the vehicle traveled a total of 7 miles (7 coming from the area under the curve over the interval (0,3)). Then the vehicle turned around and traveled back towards its initial starting point, covering 3 miles (3 coming from the area under the cuver over the interval (3, 7))"

So the car went forward 7 miles, then traveled back 3 miles. How many miles did it actually cross?

7 - 3 = 4.

(In the video we are of course scaling these values by a factor of 2 and 3, respectively). This is just an example to help you get a grasp on what is actually happening here in terms of "real world applications" and why it is structured in the way that it is.
• How could you say that the value of an integral in a function that is negative is negative? Wouldn't that be like say that the area is negative?
(1 vote)
• The ordering on these videos is really messed up. Graphical, part 2 of 2 comes before this video, which is part 1 of 2?

## Video transcript

- [Voiceover] So what we're gonna do in this video is several examples where we evaluate expressions with definite integrals and so right over here we have the definite integral from negative two to three of 2 f(x)dx plus the definite integral from 3 to seven of 3 f(x)dx and all we know about f(x) is the graph of y=f(x) from negative, from x equals negative six to x equals seven. They also give us the areas between f(x), y equals f(x) and the x axis. The negative areas show that our function is below the x axis and so given that, can we evaluate that and like always, pause the video and see if you can do it on your own. Well the first thing that my brain wants to do is I want to, I want to take these constants out of the integral because then once they're out and I'm just taking the straight up definite integrals of f(x), I can relate that to the areas over here and I know I can do that and this is a very common integration property and applies to definite and indefinite integrals but if I'm taking the integral of k f(x)dx, this is the same thing as k times the integral of f(x)dx. So let's just apply that property there which is really you're taking the scaler outside of the integral. To say this is going to be the same thing as two times the definite integral from negative two to three of f(x)dx plus three times the integral from three to seven of f(x)dx. All right. Now can we evaluate these things? So what is this going to be? The definite integral from negative two to three of f(x)dx. Well we can view that as the area between the curves y equals f(x) and the x axis, between x equals negative two and x equals three. So between x equals two and x equals three, they give us the area between y equals f(x) and the x axis. It is seven. So this thing over here is seven and then we have the integral from three to seven of f(x) so we're gonna go from three to seven and once again, well this is going to evaluate to a negative value because f(x) is below the x axis there and it's going to evaluate to negative three. So this is all going to be this is going to be two times seven so 14 plus, 14 plus three times negative three so plus negative nine and so 14 minus nine is equal to five. This is fun. Let's do more of these. All right. Okay, so here this first integral the integral from zero to five of f(x)dx, so this is pretty straightforward. This right over here, we're gonna go from zero to five. Zero to five and of f(x)dx, so we're talking about this area there which they tell us is four so that was pretty easy to evaluate and now we're gonna subtract, we're gonna subtract going from negative eight to negative four times 2f(x), well let's just take this two outside. So if we just take this two outside, then this just becomes the integral from negative eight to negative four of f(x) and so this thing, this thing right over here evaluates to five. It's this area they're talking about. So this is all going to simplify to four minus that two that we brought out, minus two times five. Times five, which is equal to let's see four minus 10 which is equal to negative six. All right. Let's do another one of these. So here I have the integral from negative seven to negative five so I'm going from negative seven to negative five which is, it's gonna be right around there so I want to find, I want to find this area right over here. So I want to find that area. That's that and then I'm gonna go from negative five to zero. So then this is going to be going from negative five to zero so it's gonna be all of that. Now there's a couple of ways you can think about doing it. You can assume I have some symmetry here and they don't tell it for sure but it looks very symmetric around x equals negative five so you could assume that this eight is split between these two regions but an easier way to do it is just to realize look, I went from negative seven to negative five and then from negative five to zero and I'm integrating the same thing. F(x)dx so this integral I can rewrite as the integral from negative seven so negative seven all the way to zero of f(x), f(x)dx and so really that's just going to be the net area between negative seven and zero and so we have the positive eight there so this is going to be equal to the positive eight and then we have the negative one there so minus one which is equal to seven.