What's the point in splitting the integral up? Surely one could just take the integral all the way from -3 to 4 and 8 at once? I've previously learned that if you want to know the total area between the curve and the x-axis, you have to split it up and take the absolute value if any of the area is below the x-axis, but is splitting it up really necessary in this case?

It depends on your purpose for finding the area. Do you need area under the x-axis treated as a positive value or a negative value? For example, if you are going to use the integral to know how much tiling a complex pattern will need, then you need area under the x-axis to count as a positive area. But if your integral represents revenue, then the negative area would be a loss and would need to stay negative. So, it just depends on what you're trying to accomplish. Thus, if you need areas under the x-axis to be negative, you don't really need to break up the integral. If you need the area under the x-axis to count as a positive area, then you need to break it up. Example: ∫ sin x dx over x = −π to π This integral obviously equals 0, if areas under the x-axis are counted as negative. But if they are counted as positive, then you would have to break them up at x=0 and change the sign for the negative region. Thus, you would need this: ∫ { sin x dx over x = 0 to π } − ∫ { sin x dx over x = −π to 0} And, obviously, that integral is equal to 4

why did sal subtracted area above x axis from area below x axis. If these are both areas why didn't he add them up

Whether to add them or subtract them depends on what purpose you're computing the integral. Does it make physical sense for the integral to represent a negative area? If so, the area beneath the x-axis is a negative area and needs subtracting. However, if it does not make physical sense for there to be a negative area, then you break up the integral and add the areas beneath the x-axis. For example, if the integral represented the profit a company made, then a negative area might represent the losses sustained. That area would need to be subtracted from the positive areas to find the actual profit. But if the integral represented the area of a complicated floor plan, then areas beneath the x-axis would add to the overall area and so the integral would have to be broken up in such a way that those areas could be added instead of subtracted. If it is not specified, _as is often the case in a calculus class_, whether the areas beneath the x-axis genuinely count as a negative area then you should assume that they are negative areas that need to be subtracted out.

The definite integrals between 4 &6 and 6&8 cancel each other out. Couldn't he have just figured those out in parts and added it to the -3.5 rather than going back and re-calculating what the whole yellow triangle would be?

Yes of course, Sal just showed how to solve it in general.

Is there a specific reason why Sal chose G(4) and G(8) at 0:38 ?

So it seems. It's a clever choice because it shows us that the chunk of net area between x=4 and x=8 is actually zero, not affecting the net area ''gathered'' in the interval [-3,4].

how to check whether a give question is integrable or not?

Just Keith has given you useful properties to look for in practice. There is a famous (and advanced) theorem, however, known as the _Riemann-Lebesgue criterion_ (or similar) which states that if `a < b` are two real numbers, and if `ƒ: [a, b] → *R*` is a bounded real-valued function on `[a, b]`, then `ƒ` is Riemann-integrable on `[a, b]` if and only if the set of points in `[a, b]` at which `ƒ` is discontinuous has Lebesgue measure zero. Informally this means that `ƒ` can only be discontinuous on a "small" set of points, in a precise sense.

I'm confused. how do you know when you're supposed to find the anti-derivative, and when you're supposed to find the area beneath the graph?

An anti-derivative is a _function_, which _can_ be used to find the area under a graph. But the specific area under a graph between two limits is always a number. So if they say, "Find the function whose derivative is 3x^2", then your answer would be "x^3 + C". If they say, "Give an expression that represents the area from x = 2 to x = k under the graph of y = 3x^2", then your answer would be k^3 - 8. Finally, if they say, "Calculate the area from x = 2 to x = 3 under the graph of y = 3x^2", then your answer would be 27 - 8 = 19.

Main content

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 6

Lesson 6: Applying properties of definite integrals

Worked example: Breaking up the integral's interval

Name: Worked example: Breaking up the integral's interval
Uploaded: 2014-05-30T17:11:18Z
Description: Finding a definite integral by breaking it down to smaller intervals that are adjacent to each other.

AP.CALC:

FUN‑6 (EU)

FUN‑6.A (LO)

FUN‑6.A.1 (EK)

FUN‑6.A.2 (EK)

Google Classroom

Finding a definite integral by breaking it down to smaller intervals that are adjacent to each other. Created by Sal Khan.

Want to join the conversation?

Sort by:

Magnus Lund
Posted 9 years ago. Direct link to Magnus Lund's post “What's the point in split...”
What's the point in splitting the integral up? Surely one could just take the integral all the way from -3 to 4 and 8 at once? I've previously learned that if you want to know the total area between the curve and the x-axis, you have to split it up and take the absolute value if any of the area is below the x-axis, but is splitting it up really necessary in this case?
Button navigates to signup pageButton navigates to signup page
(20 votes)
Answer
- Just Keith
  Posted 9 years ago. Direct link to Just Keith's post “It depends on your purpos...”
  It depends on your purpose for finding the area. Do you need area under the x-axis treated as a positive value or a negative value?
  
  For example, if you are going to use the integral to know how much tiling a complex pattern will need, then you need area under the x-axis to count as a positive area. But if your integral represents revenue, then the negative area would be a loss and would need to stay negative.
  
  So, it just depends on what you're trying to accomplish.
  
  Thus, if you need areas under the x-axis to be negative, you don't really need to break up the integral. If you need the area under the x-axis to count as a positive area, then you need to break it up.
  
  Example: ∫ sin x dx over x = −π to π
  This integral obviously equals 0, if areas under the x-axis are counted as negative. But if they are counted as positive, then you would have to break them up at x=0 and change the sign for the negative region. Thus, you would need this:
  
  ∫ { sin x dx over x = 0 to π } − ∫ { sin x dx over x = −π to 0}
  And, obviously, that integral is equal to 4
  Button navigates to signup page
  (47 votes)
Saleha Nouman Butt
Posted 9 years ago. Direct link to Saleha Nouman Butt's post “why did sal subtracted ar...”
why did sal subtracted area above x axis from area below x axis. If these are both areas why didn't he add them up
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Just Keith
  Posted 9 years ago. Direct link to Just Keith's post “Whether to add them or su...”
  Whether to add them or subtract them depends on what purpose you're computing the integral. Does it make physical sense for the integral to represent a negative area? If so, the area beneath the x-axis is a negative area and needs subtracting. However, if it does not make physical sense for there to be a negative area, then you break up the integral and add the areas beneath the x-axis.
  
  For example, if the integral represented the profit a company made, then a negative area might represent the losses sustained. That area would need to be subtracted from the positive areas to find the actual profit.
  
  But if the integral represented the area of a complicated floor plan, then areas beneath the x-axis would add to the overall area and so the integral would have to be broken up in such a way that those areas could be added instead of subtracted.
  
  If it is not specified, as is often the case in a calculus class, whether the areas beneath the x-axis genuinely count as a negative area then you should assume that they are negative areas that need to be subtracted out.
  Comment on Just Keith's post “Whether to add them or su...”
  (28 votes)
JennyB
Posted 9 years ago. Direct link to JennyB's post “The definite integrals be...”
The definite integrals between 4 &6 and 6&8 cancel each other out. Couldn't he have just figured those out in parts and added it to the -3.5 rather than going back and re-calculating what the whole yellow triangle would be?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- matthias.estner
  Posted 9 years ago. Direct link to matthias.estner's post “Yes of course, Sal just s...”
  Yes of course, Sal just showed how to solve it in general.
  Button navigates to signup page
  (7 votes)
hussein alawami
Posted 8 years ago. Direct link to hussein alawami's post “Is there a specific reaso...”
Is there a specific reason why Sal chose G(4) and G(8) at
0:38
?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Nabla ∇
  Posted 8 years ago. Direct link to Nabla ∇'s post “So it seems. It's a clev...”
  So it seems.
  
  It's a clever choice because it shows us that the chunk of net area between x=4 and x=8 is actually zero, not affecting the net area ''gathered'' in the interval [-3,4].
  Button navigates to signup page
  (4 votes)
manjit.bhatia3
Posted 8 years ago. Direct link to manjit.bhatia3's post “how to check whether a gi...”
how to check whether a give question is integrable or not?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Qeeko
  Posted 8 years ago. Direct link to Qeeko's post “Just Keith has given you ...”
  Just Keith has given you useful properties to look for in practice.
  
  There is a famous (and advanced) theorem, however, known as the Riemann-Lebesgue criterion (or similar) which states that if a < b are two real numbers, and if ƒ: [a, b] → R is a bounded real-valued function on [a, b], then ƒ is Riemann-integrable on [a, b] if and only if the set of points in [a, b] at which ƒ is discontinuous has Lebesgue measure zero. Informally this means that ƒ can only be discontinuous on a "small" set of points, in a precise sense.
  Button navigates to signup page
  (4 votes)
Cephas Simon
Posted 5 years ago. Direct link to Cephas Simon's post “how can i integrate a neg...”
how can i integrate a negative one power of (1+x)
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
AbananaK1🍌
Posted 6 years ago. Direct link to AbananaK1🍌's post “I've finished a few pract...”
I've finished a few practice sets now where you find the area between the function and the x-axis to find the value of the definite integral, but what I want to know is: why is the definite integral equal to that area? What does the area between the function and the x-axis have to do with the anti-derivative?

UPDATE
My question never got answered, but as I continued I eventually came to this video, which pretty much answered my question. So, if anyone else was wondering, here's the video:
https://www.khanacademy.org/video/proof-of-fundamental-theorem-of-calculus
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
Ajeet Dhaliwal
Posted 8 years ago. Direct link to Ajeet Dhaliwal's post “how does this topic relat...”
how does this topic relate to the fundamental theorems of calculus?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
Rob Eukovich
Posted 3 years ago. Direct link to Rob Eukovich's post “If there were no graph, h...”
If there were no graph, how would you know you would need to split the integral from -3 to 0 and then from 0 to 4 instead of simply doing -3 to 4?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- JPOgle ✝
  Posted 4 months ago. Direct link to JPOgle ✝'s post “The only reason splitting...”
  The only reason splitting the interval was useful was that this allowed Sal to calculate the area geometrically. There is no rule that makes you split the integral there. It would probably be better not to split it if he was not given the graph.
  Button navigates to signup page
  (1 vote)
Olivia
Posted 6 years ago. Direct link to Olivia's post “I'm confused. how do you ...”
I'm confused. how do you know when you're supposed to find the anti-derivative, and when you're supposed to find the area beneath the graph?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- robshowsides
  Posted 6 years ago. Direct link to robshowsides's post “An anti-derivative is a _...”
  An anti-derivative is a function, which can be used to find the area under a graph. But the specific area under a graph between two limits is always a number.
  So if they say, "Find the function whose derivative is 3x^2", then your answer would be "x^3 + C".
  If they say, "Give an expression that represents the area from x = 2 to x = k under the graph of y = 3x^2", then your answer would be k^3 - 8.
  Finally, if they say, "Calculate the area from x = 2 to x = 3 under the graph of y = 3x^2", then your answer would be 27 - 8 = 19.
  Button navigates to signup page
  (2 votes)

Video transcript

So this right over here is the graph of the function g of t. It is a function of t, and let's define a new function. Let's call it capital G of x, and it's equal to the definite integral between t is equal to negative 3 and t is equal to x. Of our lowercase g of t, g of t, dt. So, given how we have just, given how we have just defined capital G of x, let's see if we can evaluate some val, let's see if we can find some values. So, let's evaluate the function capital G. At x is equal to 4, and let's also evaluate the function capital G at x is equal to 8. And I encourage you to pause the video right now and try to think about these on your own, and then we can work through them together. So let's tackle capital G of 4 first. So this is going to be, this is going to be, well, if x is equal to 4. This, this top boundary is going to be 4, so this is going to be the definite integral of g of t between t is equal to negative 3, and t is equal to 4. g of t, g of t, d t. Now what's that going to be equal to? Well, let's see, let's look at this graph here. So, we have negative 3, this t is equal to negative 3, which is right over here. T is equal to negative 3. And we're going to go all the way to is equal to 4. Let me circle that in orange. All the way to t is equal to 4 which is right over here. So this is. One way to think about this is this is going to be the area above t-axis and below the graph of G. So, it's going to be this area, right over here, that is above the t-axis and below the graph of G of T, but, we're not going to add to it this area because this area over here. I'll shade it in yellow. This area I'm shading in yellow over here, this is going to be negative. Why's it going to be negative? Because we want the area that is above t and below g. This is the other way around. This is below the t axis and above the graph of g. So one way to think about it is, we could split it up. So, we could, actually let me just clear this out so we have more space. So, this, right over here, is going to be equal to the integral. I'll do that in this purple color. The integral from t equals negative 3 to 0, of g of dt +. I'll do this in the other color. Plus the interval from 0, t equals 0 to t is equal to 4 of g of t, dt. Now what are each of these going to be? Well this is just a triangle where the base is 3. The base has length 3. The height is length 3. So, it's going to be 3 times 3, which is 9 times one half. Because 3 times 3 would give us the area of this entire square. But this triangle is half of that. So it's, this right over here, is going to be 4.5. This is going to be 4.5. And then, what's this area in yellow? Well, let's see, we have a triangle, it's base is, or it's width here is 4, it's height is 4, 4 times 4 is 16, which would be area of this entire square. We take half of that, it's 8. Now we don't just add it to it, because, once again, this is going to be negative area. The graph over here is below, below the T axis. So this, we would say, we would, this integral is going to evaluate to negative 8. Once again, why is it negative 8? Cause the graph is below, below the T axis. And so what do we get? We get g(4) which is this area essentially minus this area. 4.5 minus 8 is going to be. Let's see 4 minus 8 is negative 4 out of .5 which is negative 3.5. Negative 3.5. Now let's try to figure out what g of 8 is. So g of 8. And if you couldn't figure it out the 1st time around, try to pause the video again, and now that we've figured out g of 4, try to figure what g of 8 is. Well g of 8 is, 1 way to think about it, it's going to be that. Minus that area and then we're going to add, then we're going to figure out the area and then we're going to have two more areas to think about. We're gonna go all the way to 8. So actually make, draw the line there. So we're gonna have to think about, we're going to think about this whole area now. This whole area now. And then we're gonna think about this one. So we could write this is going to be equal to the integral between. Actually let me do that purple color. So it's going to be this, this integral. Which is the integral between negative, t equals negative 3 and 0. G of t, dt. Plus this entire yellow region right now. The part that we looked at before plus this part over here. So, I'll just write that as, plus the definated role between t is equal to 0 and 6 of g of t dt. And then finally. Plus the definite integral between T is equal to 6 and T is equal to 8 g(t) dt. Now we already know that this first one evaluates to 4.5, so that one is 4.5. Now what's this one going to be? We have a triangle. Whose base is length 6, height is 4, 6 times 4 is 24 times one half is going to get us 12. So this is going to be, that over there is twelve and then finally what is this area? Oh and we have to be careful. This is below the t axis and above the graph. So this is going to be a negative 12. And then finally, we have this area, which is once again going to be a positive area. It's below the graph and above the t axis. And so let's see, 2 times 4 is 8 times 0.5 is 4. So we're going to have. Plus 4. Plus 4. And so what do we have? We end up with 4.5 plus 4 is 8.5 minus 12. So this is going to be equal to. This is going to be equal to 8.8 minus 12 would be negative 4 plus 5. It is negative 3.5 again. Now why did these two things end up being the same? Well think about what happened here. When we went from capital G of 4 to capital G of 8. We subtracted, we subtracted this value right over here and we added this value right over here and you see that these two triangles have the same areas. So we subtracted and added the same amount to get the same exact value.