help needed with this antiderivative?
605867
(x^2)/2 - 3(x^(3/2))/(3/2) + c = (x^2)/2 - 2x^(3/2) + c
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
1166133
Multiply both sides by R+r\nI(R+r) = E\nDistribute the I\nIR + Rr = E\nSubtract IR form each side\nRr = E - IR\nDivide both sides by R\nr = (E-IR)/R
969305
This initial value problem is surprisingly simple, since niether y nor any derivatives other than the second appear in the equation, so all that is necessary is to integrate twice:\n∫∫d²y/dx² dx dx = ∫∫6x dx dx\n∫dy/dx dx = ∫3x² + C dx\ny=x³+Cx+K\n\nNow we need merely find the constants C and K. We know that y'(0)=3(0)² + C = 0, so C=0. Further, we know that y(0) = K = 1, so K=1.\n\nThus y=x³+1\n\nb: There is exactly one such function. All antiderivatives of a function are the same up to addition of an arbitrary constant, and as we showed when solving the problem, there is only one consistent set of values for the two constants of integration introduced in this problem.
982407
dy/dx=cosx*e^y*e^sinx\ndy/e^y= cosx*e^sinx*dx\ne^-ydy= cosx*e^sinx*dx\nintegrate\ne^-y/(-1)=int. cosx*e^sinx*dx\nput sinx=t\ncosxdx=dt\n-e^-y=int.e^tdt\n        =e^t+c\n        =e^sinx+c\nput x=0,y=0\n-1=1+c\nc=-2\n-e^-y=e^sinx-2
79944
The above is the derivative.\nIntegration gives x + (1/4)x^4 + c (generic constant).
761744
use the product rule, with the first term as t and the second as (4-t)^1/2\nproduct rule says that the derivative is second term times the derivative of the first term plus first term times the derivative of the second term.  Don't forget to use the chain rule along the way.\n\nThat is\nxy would have a derivate of yx' =xy'\n\nSo here\n[(4-t)^1/2](1) + (t){[1/2(4-t)^ -1/2][-1]\n(4-t)^1/2 + -t/2(4-t)^ -1/2
1181156
It looks like that first 1 should be an i. In this case,\ni^z=exp(i*pi/2*z),\nwhere I am using the principal branch of the logarithm to define the exponential.\nThen the derivative will be \n(i*pi/2)*exp(i*pi/2*z)=\npi/2*i^(z+1)\n\nEasy.
770990
I don't understand your statement of the problem.  It sounds like it might be an exercise in integral calculus, in which you integrate the expression at hand, evaluate the resulting function at the two limits, and take the difference.
676902
Incomplete question. Are you looking for the anti-derivative of x, a, or b???\n\nput dx, da, or db at the end of the equation so we'll know.
871386
∫udv = uv - ∫vdu\n∫x^2dx/√(1-x)\nlet u = x^2, dv = dx/√(1-x)\ndu = 2xdx, v = -2√(1-x)\n∫x^2dx/√(1-x) = -2x^2√(1-x) + ∫(4x√(1-x))dx\nu = 4x, dv = √(1-x))dx\ndu = 4dx, v = -(2/3)(1-x)^(3/2)\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) + ∫(8/3)(1-x)^(3/2)dx\n∫x^2dx/√(1-x) = -2x^2√(1-x) - (8x/3)(1-x)^(3/2) - (16/15)(1-x)^(5/2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - (8x/3)(1 - x) - (16/15)(1-x)^2)\n∫x^2dx/√(1-x) = √(1-x)(2x^2 - 8x/3 - 8x^2/3 - (16/15)(1 - 2x + x^2)\n∫x^2dx/√(1-x) = (1/15)√(1-x)(30x^2 - 40x - 40x^2 - 16 + 32x + 16x^2)\n∫x^2dx/√(1-x) = -2√(1 - x)(x(3x + 4) + 8)/15
710286
the antiderivative of 4 / (t^2 + 1)  IS \n4 arctan(t) + C`
199351
I haven't worked with this in a long time and I'm not sure this will work, but try starting w/ this:\n1) Define an equation that describes the " ∞ " with polar coordinates (something like r = C1 - C2*theta);\n2) If L is lengh, delta.L = delta.theta / (2 pi) * r\nand dL = r/(2 pi) dtheta  ;\n3) substitute for r and integrate from 0 to 2pi (or the necessary for one complete " ∞ ".\n \nProbably you'll find more than one mistakes here, but I think the general idea is correct.
971390
{3XY^2M/1}-{2M/[X^-1Y^-1]^2\n+{[4X^2Y^2]/M^-1}-\n(3X^(3Y)^(3M)^-1/(XYM)^-2
856976
Extreme position! \nBecause acceleration is directly propotional to the displacement from the mean position and directed opposite to it.\n\na=-(ω^2)x
806346
using implicit differentiation:\nyou get 2y(dy/dx) - [y + x(dy/dx)] -3 = 0\nthen move 3 over and factor out (dy/dx):\n(2y-x)(dy/dx)=3 + y\ndy/dx=(3+y) / (2y-x)\nnow plug in your point (0,-1)\ndy/dx = (3+ -1)/(2(-1) - 0\ndy/dx = -1\nanswer = -1\n\n(dont forget to use your chain rule when you are implicitlly differentiating)\n\n#2\nsame thing make sure you sue multiplication rule\n[y^2 + 2xy(dx/dy)] - 2(dx/dy) + 12y^2(dx/dy) = 0\nthen move over the y^2 and factor out dx/dy\n(dx/dy)[2xy - 2 + 12y^2] = y^2\ndx/dy = y^2/ (2xy -2 +12y^2)\nPlug in 1 for y\ndx/dy = 1/(2x -2 + 12)\ndx/dy = 1/ (2x +10) or (2x + 10)^(-1)\n\nAnswer (leave in terms of x) = 1/ (2x +10) or (2x + 10)^(-1)\n\nI may have made mistakes so let me know