Benjamin753

Was wondering if anyone could tell me how I'd go about integrating: Int[(r(z))^2 dz] on the interval [1,h] where r(z) is an arbitrary function of z. help is appreciated:)

I'm not sure if this belongs here or in the homework section, but, I'm looking for an online resource of problems for a differential calculus class (derivatives, optimization, related rates, etc.). I've google'd it and found a few, but a lot are dead links or only have one or two problems, and a whole lot of explanations. I'm just looking for problems, with answers. I thought someone here might be kind enough to point me in that direction. Thanks in advance:)

I said other... they should release all the female Iraqi prisoners, (as they shouldn't have invaded the country anyway) and the Iraqi's should get to keep Ken as a very small beginning to the reparations for a hostile and unwarranted invasion of a soverign country... Hell, they should give them Tony Blair too.

Nevermind... I got it... my conjugate was wrong... oops.

Hi anybody... I have a problem. Someone (or most) should find it really easy, but I'm having trouble. Here it is. f(x)= x^(1/3) (or the cube root of x) need to find f'(x) using limit as h->0 of (f(x+h) - f(x))/h Now, I already know what the answer is: (1/(3(x^(2/3))), but how do I get that using the difference quotient (above). Obviously, my algebra skills are rusty, but here is what I did to get the wrong answer: ((x+h)^(1/3)) - (x^(1/3)) ( times the conjugate) ------------------------ h ((x+h)^(2/3)) + (x^(2/3)) ------------------------- ((x+h)^(2/3)) + (x^(2/3)) = x + h - x --------------------------- h((x+h)^(2/3)) + (x^(2/3)) = 1 ---------------------------- ((x+h)^(2/3)) + (x^(2/3)) (then sub 0 for h, because it is a limit problem) then ----> f'(x) = 1 ------------ 2(x^(2/3)) which is not correct... Anybody help me out with this one?? I would appreciate it! Thanks!