Recent content by ShaneOH

Hey guys, I'm sorry to bring the same problem up again, but my answer is lacking two entire terms and I'm not sure what I'm omitting. Problem: ∬ √(x + 4y) dxdy, where R = [0, 1] x [2, 3] Solution attempt: I would assume that, u = x + 4y du = 1 dx (since we're integrating with respect to x...

So they become variable bounds. Got it. Thanks so much for the help and patience Sourabh!

So (last question!) what you're saying is, if this problem would have been such that u = 2x + 4y, then the bounds would have been from 4 to 6? I literally just completely drop the y? Ignore it? Say, u = 2x + xy + y, and normal bounds are from 2 to 4 Then, change of bounds: u = 2(2) + 2...

Sure, so I'm confused on: Single variable substitution: Let's say you have some arbitrary integration problem from 2 to 4, of √(x^2 + x). Here, u = x^2 + x du = 2x + 1 dx dx = du/2x + 1 And so, instead of 2 to 4, the bounds become: (2)^2 + (2) = 6 (4)^2 + 4 = 20 So the...

@Sourabh. Sure thing, it just looked... wrong, but I'll take a crack at continuing. @daveb: I was under the impression I needed u-substitution for more than one value under a square root? Or does it just become (2(x+C)^(3/2))/3)? Also: Could someone just shed some light on my change of...

Homework Statement Hey guys, so we're going over multiple integration in Calc III, and I'm having trouble with the more complex problems. ∬ √(x + 4y) dxdy, where R = [0, 1] x [2, 3] So it's 0 to 1 on the outside integral, 2 to 3 on the inside integral, of sqrt(x+4y) dxdy. Homework...