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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 4

Lesson 13: Functions defined by integrals- Worked example: Breaking up the integral's interval
- Functions defined by integrals: switched interval
- Functions defined by definite integrals (accumulation functions)
- Functions defined by integrals: challenge problem
- Functions defined by integrals challenge

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# Worked example: Breaking up the integral's interval

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

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?

- 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?(20 votes)
- 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(47 votes)

- 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(6 votes)
- 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.(28 votes)

- 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?(5 votes)
- Yes of course, Sal just showed how to solve it in general.(7 votes)

- Is there a specific reason why Sal chose G(4) and G(8) at0:38?(5 votes)
- 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].(4 votes)

- how to check whether a give question is integrable or not?(1 vote)
- 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] →`

is a bounded real-valued function on**R**`[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.(4 votes)

- 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(3 votes) - how does this topic relate to the fundamental theorems of calculus?(3 votes)
- how can i integrate a negative one power of (1+x)(2 votes)
- 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?(1 vote)
- 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.(1 vote)

- 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?(1 vote)
- 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.(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.