Integrals, limits, and derivatives???
338000
I only had time to do the first one, btu the answer is \ne-2.  Check your email for the work.
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
79903
Here is an excellent site:\nhttp://www.sosmath.com/calculus/calculus.html\n\nVery thorough explanations and examples provided.
796927
try the link below...... hope it can put an end to ur integral problems
353544
http://mathforum.org/library/drmath/sets/high_calculus.html\n\nask here...:)
1047841
By the def\n\nlim f(x + h ) - f(x)\n              /h\nPlug in (x+h) into f\nlim (x+h)^2 + 5(x +h) - (x^2 + 5x)\n                        /h\nFoil\nlim x^2 + 2xh + h^2 + 5x + 5h - x^2 - 5x\n                          /h\n\nCancel like terms\nlim 2xh + h^2 + 5h\n           /h\n\n\nReduce everything by h\nlim 2x + h + 5\n\nTake the limit and\nPlug in 0 for h\n2x + 5
794871
assuming 0 and 3 are limits of integration:\nv(t) = dx/dt = t^2 - 2t\nx = (1/3)t^3 - t^2 + C\nx = (1/3)3^3 - 3^2 - 0 + 0\nx = 9 - 9 = 0\n\nIf the limits are 0 and 2, \nx = 8/3 - 4 = -4/3\n\nGoing backwards,\n(1/3)t^3 - t^2 = 8/3\nt^3 - 3t^2 -8 = 0\nt ≈ 3.612887865
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
424650
Not really enough information to help, so try this site:\n\nhttp://www.blc.edu/fac/rbuelow/calc/ex4-6.html\n\nit's the second problem down the page
2024
Case 1: f(x) := x-5*y; (y is parameter)\n\nL{f(x)} = int(exp(-s*x)*f(x),x=0..infinity);\nL{f(x)} = int(exp(-s*x)*(x-5*y),x=0..infinity);\nL{f(x)} = 1/s^2-5*y/s;\n\nCase 2: f(y) := x-5*y; (x is parameter)\n\nL{f(y)} = int(exp(-s*y)*f(y),y=0..infinity);\nL{f(y)} = int(exp(-s*y)*(x-5*y),y=0..infinity);\nL{f(y)} = x/s-5/s^2;
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/
101992
Here are some sites that have saved my behind.  :)\n\nhttp://integrals.wolfram.com/index.jsp\n\nhttp://www.calc101.com/webMathematica/integrals.jsp\n\nhttp://mcraefamily.com/MathHelp/CalculusTableOfIntegrals.htm
945645
z = y - 2x\nwhen z= 0 , y=2x\nwhwn y=2, x=1\n\nSo, limits for x are 0 to 1, \nlimits for y are 2 to 2x\nlimits for z are 0 to y-2x\n\nSo ,you can write your integral as\n\nint(x=0 to x=1) int(y=2 to y=2x) int(z=0 to z=y-2x) [f(x,y,z) dxdydz]\n\nYou can change the limits and order of variables with availabel theorems.
772619
integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
887032
This is a fundamental theorem of calculus problem.\n\nThe derivative of the integral of t * sqrt (t^3 + 2t) dt from -1 to x is\n\nx * sqrt (x^3 + 2x)
972569
By definition of the derivative\ng'(x) = lim h=>0 [g(x + h) - g(x)] /h =  lim h=>0 [3xh + h^2]/h = \nlim h=>0 [3x + h] = 3x
574666
int((x-2)*(x-5)^(1/2)*dx)\n\nUse u sub, let:\nu=x-5\ndu=dx\nx-2=u+3\n\nResub:\n\nint((u+3)*sqrt(u)*du)\n\nDistribute:\n\nint((u^3/2 + 3sqrt(u)*du))\n\nIntegrate:\n\n0.4*u^(5/2) + 2*u^(3/2) + c\n\nResub u for x:\n\n0.4*(x-5)^(5/2) + 2*(x-5)^(3/2) + C