Integrating for area of intersection

[Rumplestiltskin]

How does it work that you can subtract y₂ from y₁ and integrate the product within defined limits for the area of their intersection (within those limits)?
Maybe that's not the right terminology - you arrive at the area for the region bounded by both functions.

Is it just the same in practice as subtracting the integral of y₂ from the integral of y₁?

 

[462chevelle]

I'm not 100% what you're asking, but it sounds to me like you are trying to find the area bound by the intersection of two curves. One thing you may have to consider is if they only intersect at two points within those limits. If they only intersect at two points within your limits, you can simply just do the integral of y2 minus the integral of y1. With y2 being the upper function on the graph.

 

[Rumplestiltskin]

I'm not 100% what you're asking, but it sounds to me like you are trying to find the area bound by the intersection of two curves. One thing you may have to consider is if they only intersect at two points within those limits. If they only intersect at two points within your limits, you can simply just do the integral of y2 minus the integral of y1. With y2 being the upper function on the graph.

How's that?

 

[Mark44]

How does it work that you can subtract y₂ from y₁ and integrate the product within defined limits for the area of their intersection (within those limits)?
Maybe that's not the right terminology - you arrive at the area for the region bounded by both functions.

Is it just the same in practice as subtracting the integral of y₂ from the integral of y₁?

Let's call the line f(x) and the curve g(x). Assuming that the line and curve intersect at a and b, the area of the region in blue is $\int_a^b (f(x) - g(x)) dx$. This is exactly the same as $\int_a^b f(x) dx - \int_a^b g(x)dx$, due to the linearity properties of the definite integral.

The product that you refer to is the area of the typical area element, $\Delta A = (f(x) - g(x)\Delta x$. When you integrate, you are essentially summing all of these incremental area elements to get the total area.

How's that?

 

[Rumplestiltskin]

Let's call the line f(x) and the curve g(x). Assuming that the line and curve intersect at a and b, the area of the region in blue is $\int_a^b (f(x) - g(x)) dx$. This is exactly the same as $\int_a^b f(x) dx - \int_a^b g(x)dx$, due to the linearity properties of the definite integral.

Due to the linearity properties of the definite integral?

 

[Mark44]

Due to the linearity properties of the definite integral?

$\int_a^b (f(x) + g(x))~dx = \int_a^b f(x)~dx + \int_a^b g(x)~dx$
and
$\int_a^b k \cdot f(x)~dx = k\int_a^b f(x)~dx$ where k is any constant
The same properties hokl for indefinite integrals, as well.

 

[Rumplestiltskin]

$\int_a^b (f(x) + g(x))~dx = \int_a^b f(x)~dx + \int_a^b g(x)~dx$
and
$\int_a^b k \cdot f(x)~dx = k\int_a^b f(x)~dx$ where k is any constant
The same properties hokl for indefinite integrals, as well.

LOL. I take it I'm punching well above my weight again.

 

[462chevelle]

I think you aren't as confused as you think. He us just showing properties of the integral in a general sense. That way of writing might be intimidating if you don't have any formal knowledge on the subject.