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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?
• 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 ?
• 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?
(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] → 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.
• how does this topic relate to the fundamental theorems of calculus?
• how can i integrate a negative one power of (1+x)
• 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: