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yeah that's right so far, let me make sure I have the answer though...

yeah it sure does...
think about this: what is x^2 when x=0 ?
what is 8-2x when x=0 ?

Well x^2 would be 0 and 8-2(0) would be 4

so 8-2x should be on the top, yes?

Oh so it should be (8-2x)-(x2) instead?

...hence the negative answer

Okay let me try this and see if it is better

should be :D

dang I wish I knew all this like you ha would make this extensive homework easier

There is only one more question I had i doubt i can explain it here though. Let me try quick.

I only see two shaded regions between these graphs:
between -1,0 and 0,1

Our drawing is from -2 and 0, 0 to 3 ish? but some is above the x/3 line and some below

|dw:1328632999874:dw|

Not the best drawing but it wants the regions

click the link for a better image of the graph

The shaded region I drew. I think thats the region between graphs?

Well i think its bounded between the line and x=3

so he wants the area from the line at x=3 so it now has bounds i guess

then your intervals are\[[-1,0][0,1][1,3]\]

are u sure its not -2,0?

my book has the points (-2,2/3) as one point and (3,6) as another

if you want the region between the graphs then\[\frac{-2}{3}\neq\frac{(-2)^3}3\]so no

at least that's not a point of intersection

(-2,-2/3)

waitt

i feel stupid sorry. the line is x^3/3 -x

and x/3

ah...
it happens

should make more sense

Well the 3rd shaded region the lines dont intercept

Its just from the point of (3,6) straight down to (3,1)

me too, got class
you should be able to set up the integral now though, good luck!

Thanks, good luck

Is c a constant or a variable?