How to determine how many areas there are between two functions

[hatelove]

I was given a problem with two functions and two x-values for boundaries, so I found the points of intersection (there were two) and attempted to find the area between those functions, but I didn't get to finish. In any case, I would have gotten it wrong, because when graph the two functions and then look at the boundaries, there are 3 separate areas that needed to be added up. I just thought it was another problem with a parabolic curve and a line going through it, and there was only one area in between them, but it was asking me to also include the areas that were not in both functions, i.e.

Without a graphing utility and short of graphing both functions very accurately to tell this, is there any other way to see that this is what the question was asking for?

 

[SuperSonic4]

I was given a problem with two functions and two x-values for boundaries, so I found the points of intersection (there were two) and attempted to find the area between those functions, but I didn't get to finish. In any case, I would have gotten it wrong, because when graph the two functions and then look at the boundaries, there are 3 separate areas that needed to be added up. I just thought it was another problem with a parabolic curve and a line going through it, and there was only one area in between them, but it was asking me to also include the areas that were not in both functions, i.e.

<snip>

Without a graphing utility and short of graphing both functions very accurately to tell this, is there any other way to see that this is what the question was asking for?

What's the actual question? I'm trying to think it through but it'll be easier with a real example.

 

[hatelove]

I don't remember it, but I will try to find one or some up with my own, hold on

 

[hatelove]

y = x + 10 and y = x^2 + 5, find the area between x = -7 and x = 6

 

[pickslides]

y = x + 10 and y = x^2 + 5, find the area between x = -7 and x = 6

First step is find where they intersect.

 

[hatelove]

Well I know that, set them equal to each other and it's a couple of irrational numbers (sorry I didn't come up with a cleaner problem; I just picked random numbers instead of multiplying factors), and then I just assume that it's one area and use the integral formula for finding the area of that. I have no idea it's 3 different areas, and would not be able to tell without looking at the exact graph and shading in areas.

In my mind, I am thinking:

 

[pickslides]

Well let's call these solutions a and b, where a < b. Now graphing the functions you will see one is greater than the other between the intervals (-7 and a), (a,b) & (b,6). Integrate the difference of the functions between these intervals subtracting the lesser function from the greater on each individual interval.

 

[SuperSonic4]

edit: too slow :(

y = x + 10 and y = x^2 + 5, find the area between x = -7 and x = 6

They meet at: $x = \dfrac{1\pm\sqrt{21}}{2}$.

Let $\alpha = \dfrac{1}{2}(1-\sqrt{21}) \text{ and } \beta = \dfrac{1}{2}(1+\sqrt{21})$

Let $f(x) = x+10$ and $g(x) = x^2+5$. Work out the points at the lower and upper bounds so we know which order to subtract the integral from (so we don't have a negative answer)

• $f(-7) = ?$
• $ f(6) = ?$
• $g(-7) = ?$
• $g(6) = ?$

You want to find the area under g(x) between the lower bound (x=-7) and the negative intersection to the x-axis. If you do the same for f(x) between the same two limits then you can subtract the area under f(x) from the area under g(x).

Spoiler

$\displaystyle \int^{-7}_{\alpha} g(x)dx - \int^{-7}_{\alpha}f(x)dx$

i.e. $\displaystyle \int^{-7}_{\alpha} (x^2+5)dx - \int^{-7}_{\alpha}(x+10)dx$

Then you do it between the points of intersection as you know how

You then do the same thing for the area between the positive intersection and the upper bound

Spoiler

$\displaystyle \int^{6}_{\beta} g(x)dx - \int^{6}_{\beta}f(x)dx$

i.e. $\displaystyle \int^{6}_{\beta} (x^2+5)dx - \int^{6}_{\beta}(x+10)dx$

Then you add all three values to get the final answer.