Undergrad Why split interval here? Area under a curve

[opus]

Please see attached image.
When we want to find the area under a curve, we can use the formula
$A = \int_a^b\left|g(x)-f(x)\right|dx$ where g(x) is greater than f(x) and both are continuous over the closed interval $[a,b]$
My text, as seen in the picture, described the area under the curve as a complex region, and says we need to split it up.
What's the reasoning behind this? Does it have something to do with the bounds?

 

[opus]

I may have noticed a possible reason. On [0,1) $g(x) > f(x)$ and on (1,2] $f(x) > g(x)$. So I don't think we could use the given formula in this case because a function would need to be greater than the other over the entire interval.

 

[PeroK]

I may have noticed a possible reason. On [0,1) $g(x) > f(x)$ and on (1,2] $f(x) > g(x)$. So I don't think we could use the given formula in this case because a function would need to be greater than the other over the entire interval.

The region is entirely above the x-axis, so there is no need to split the integral to account of regions of "positive" and "negative" area. But, the function defining the region changes at $x=1$. So, you need to integrate one function from 0 to 1 and another function from 1 to 2.

Note that the region is not the difference of two functions.

 

[opus]

The region is entirely above the x-axis, so there is no need to split the integral to account of regions of "positive" and "negative" area.

I haven't gotten to such a problem but I'll keep this in mind when I do.

But, the function defining the region changes at x=1x=1. So, you need to integrate one function from 0 to 1 and another function from 1 to 2.

So on [0,1), the function defining the region is the orange one, and from (1,2] it's the blue?
So if I have something like the graph in this new image, which appears to have an area defined by two functions throughout, it would be acceptable to have a single integration rather than two?

 

[opus]

Note that the region is not the difference of two functions.

Thanks for pointing that out. The given formula in post 1 has a subtraction of two areas, but it makes sense that this would be an addition of the two since we're splitting it up.

 

[PeroK]

I haven't gotten to such a problem but I'll keep this in mind when I do.So on [0,1), the function defining the region is the orange one, and from (1,2] it's the blue?
So if I have something like the graph in this new image, which appears to have an area defined by two functions throughout, it would be acceptable to have a single integration rather than two?

That area is the region between two functions!

 

[opus]

That area is the region between two functions!

Yes, but you're saying since, a part of the area is strictly defined by one function, and the other part is strictly defined by another, it needs to be separated into a manner where we take the integral over the entire domain covered by one function, the integral of the rest of the domain covered by the other, and then add them up. Correct?

 

[PeroK]

Yes, but you're saying since, a part of the area is strictly defined by one function, and the other part is strictly defined by another, it needs to be separated into a manner where we take the integral over the entire domain covered by one function, the integral of the rest of the domain covered by the other, and then add them up. Correct?

Your second example is the area between two functions.

You need to be able to identify what you are integrating. For example, if I asked you to calculate the area of a triangle and square joined together, you would have to calculate the area of the triangle and the area of the square and add them together. There's nothing except your own logical thinking that will tell you that!

The first example you posted was the area between the x-axis and complicated function that was part parabola, before becoming a straight line. Like the triangle and square, you need to calculate those two parts separately. Again, you have to see that for yourself.

The second example you posted is the area between two curves. You can calculate that by integrating the difference of the two functions. Again, you have to see that for yourself.

 

[opus]

Loud and clear. Thanks!