Recent content by Casquibaldo

yes 73/6 :) thanks a lot everyone. The patience is definitely appreciated! (it's the same as ∫[1, 4] (√x -(-x))dx )

this has become very clear to me, thank you :) I think it would be something like ∫[-4, -1] (4-(-y))dy + ∫[-1, 1] (4-(1))dy + ∫[1, 2] (4-(y^2))dy

I think this is what you mean

Oh wait is is rather ∫(sqrt(x))dx from 0 to 4 - ∫(-x)dx from 0 to 4 which is 40/3? or ∫(sqrt(x))dx from 0 to 4 + ∫(-x)dx from 0 to 4 which is -8/3? or is it just neither and I'm a bit confused

would that be from -4 to 2? That means it would be -74/3 (which I think makes sense because most of the graph is negative

Ok so now that I know what the integrand is ∫(y^2 +y)dy how do I figure out from where to where it is? (like a and b)

oh I see what you mean- thanks And rest assured, I copied it just how it shows on my handout.

oh ok so x=-y but how do I go about getting the points of intersection and from which side would I integrate it since it is all sideways ?

I would think so too, but he said something about it being respect to y, and I probably didn't take that class and it might mean that I have to switch something, I don't know.

Homework Statement Set up sums of integrals that can be used to find the area of the region bounded by the graphs of the equations by integrating with respect to y. y= sqrt(x) y=-x x=1 x=4 Homework Equations ∫[f(x) - g(x)] dxThe Attempt at a Solution I did: ∫ (from 1 to 4) [sqrt(x) + x] dx...