anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
can someone help
TuringTest
  • 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\[R=\int_{0}^{x_o}g-fdx+\int_{x_o}^{6}f-gdx\]
anonymous
  • 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?

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TuringTest
  • TuringTest
whoa sorry, misread just\[R=\int_{0}^{x_o}fdx+\int_{x_o}^{6}gdx\]
anonymous
  • anonymous
i get what you are saying their but i am a little confuse to how to know what integral to set up
anonymous
  • anonymous
btw thanks for the help
TuringTest
  • 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\[R=R_1+R_2=\int_{0}^{4}fdx+\int_{4}^{6}gdx\]
TuringTest
  • TuringTest
|dw:1329270195071:dw|
anonymous
  • 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
TuringTest
  • TuringTest
g>f for 0
TuringTest
  • TuringTest
\[\int_{0}^{4}fdx\]is this area:|dw:1329270801204:dw|
TuringTest
  • TuringTest
\[\int_{4}^{6}gdx\]would be this|dw:1329270875759:dw|
TuringTest
  • TuringTest
\[\int_{0}^{4}g-fdx\]would be this|dw:1329270925750:dw|it is important to think of the shape of each area...
anonymous
  • anonymous
Ok great explanations that you!!!
TuringTest
  • TuringTest
welcome :D
anonymous
  • anonymous
I will definately be asking more questions ....hope you can help me...lol i am sure you can
TuringTest
  • TuringTest
hope so :)
anonymous
  • anonymous
let me find some more

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