help with integration q pls D: (1 Viewer)

[Rhanoct]

Find the area of the region between the graphs of y=2x-x^(2) and 3y=x^(2) - 4x +6

Wasted like an hour on it >.< I found the pts of intersection as 1, 3/2 but when i try integrate something goes wrong :S answer is 1/36 sq. units

ta! <3

 

[watatank]

try finding the points of intersection again. i'm, fair sure they are x = 1/2, 1.

 

[Rhanoct]

will do, wasn't really confident on the points.

just a quick q: when i'm finding the pts of int it's 2x-x^(2) = [x^(2) - 4x +6]/3 right?

 

[izzah1]

Integral sign - 3/2 to 1

2x - x^2 - ((x^2) - 4x +6)/3)
=
(6x - 3x^2 - x^2 +4x - 6)/3 --- by putting it all over 1 denominator

=

(-4x^2 + 10x -6)/3 --- collecting like terms

now you can divide all the terms by 3 so
=
-4x^2/3 + 10x/3 - 2

Now you integrated it

becomes -4x^3/9 + 5X^2/3 - 2x

then just sub in your values 3/2 and 1

and it ends up being 1/36 sq

Hope that helps

izzah

 

[izzah1]

Your points of intersection are correct, dw about them.

 

[Rhanoct]

thanks guys i didn't do the g(x) - f(x) part properly, grrrrr

 

[izzah1]

haha no worries, gl with the rest of the work, try to get some sleep though its almost 12