## anonymous 4 years ago the functions f and g are given f=sqrt(x) and g = 6-x. Let R be the region bounded by the x-axis and the graphs of f and g. Find the area of R

1. anonymous

can someone help

2. TuringTest

first step is to get an idea of the shape of the area:|dw:1329269119785:dw|it looks like we will have to do two separate integrals, because we can see that before the line x=xo we have the area as g-f, and after given by f-g. xo occurs when the functions intersect, so we can find it by solving$f(x)=g(x)$for x. our integral to find R will then be$R=\int_{0}^{x_o}g-fdx+\int_{x_o}^{6}f-gdx$

3. anonymous

i found the intersection point between the two curve and its (4,2) so i integrated from 0 to 4 and did f - g why is that wrong?

4. TuringTest

whoa sorry, misread just$R=\int_{0}^{x_o}fdx+\int_{x_o}^{6}gdx$

5. anonymous

i get what you are saying their but i am a little confuse to how to know what integral to set up

6. anonymous

btw thanks for the help

7. TuringTest

you found that the intersection point is at x=4|dw:1329269909705:dw|the first area is under f so it is$R_1=\int_{0}^{4}fdx$the second area is under g so it is$R_2=\int_{4}^{6}gdx$the total area is then$R=R_1+R_2=\int_{0}^{4}fdx+\int_{4}^{6}gdx$

8. TuringTest

|dw:1329270195071:dw|

9. anonymous

I guess its wrong to why is it wrong to integrate for 0 to 6 of f - g...your pic is very good i get that...but i dont get why you can just integrate from 0 to 6

10. TuringTest

g>f for 0<x<4, so f-g would be negative. we want a positive area Imagine there were no g. You can see that R1 would be given by the integral of f from 0 to 4 either way, so excluding g on that interval will give us the right answer f-g for 4<x<6 would be the black area on the following graph area:|dw:1329270527650:dw| so you can see that is not right either.

11. TuringTest

$\int_{0}^{4}fdx$is this area:|dw:1329270801204:dw|

12. TuringTest

$\int_{4}^{6}gdx$would be this|dw:1329270875759:dw|

13. TuringTest

$\int_{0}^{4}g-fdx$would be this|dw:1329270925750:dw|it is important to think of the shape of each area...

14. anonymous

Ok great explanations that you!!!

15. TuringTest

welcome :D

16. anonymous

I will definately be asking more questions ....hope you can help me...lol i am sure you can

17. TuringTest

hope so :)

18. anonymous

let me find some more