# Thread: Areas between curves - Which is on top?

1. ## Areas between curves - Which is on top?

I seem to be getting in a muddle on finding which function goes on top.

Can someone help

I understood it to be the one which has the largest y output when making the points of intersection the argument

BUT it seems to change on some functions

Thanks

2. Originally Posted by 200001
I seem to be getting in a muddle on finding which function goes on top.

Can someone help

I understood it to be the one which has the largest y output when making the points of intersection the argument

BUT it seems to change on some functions

Thanks

You have to solve inequalities, of course: after you know the integration interval, you have to solve $f(x)>g(x)$ for the two functions.

Tonio

3. Originally Posted by 200001
I seem to be getting in a muddle on finding which function goes on top.

Can someone help

I understood it to be the one which has the largest y output when making the points of intersection the argument

BUT it seems to change on some functions

Thanks
If $f(x)\geq{g}(x)$ for all $x\in[a,b]$, then $\int_a^b[f(x)-g(x)]dx$ represents the area between the two curves $f$ and $g$ on $[a,b]$.

4. Thanks
and if its not then you have to integrate seperately and subtract afterwards?

5. Originally Posted by 200001
Thanks
and if its not then you have to integrate seperately and subtract afterwards?
No, if f(x)> g(x) on parts of the interval and g(x)> f(x) for other parts, then you will need to integrate (f(x)- g(x)) for the parts where f(x)> g(x), integrate (g(x)- f(x)) for the parts where g(x)> f(x), and add. The point is that "area" is positive and so you must always be integrating a positive function. If f(x)> g(x), then f(x)- g(x)> 0. If g(x)> f(x), then g(x)- f(x)> 0.

6. Awesome, that makes sense