integral question...?
958105
Hi,\n\nThis not just ...sqrt(pi) as one respondent thought.\n\nWhat you need are two very useful tools for integration, part one is solved by changing variables and part two is solved with integration by parts.\n\n1. let x= (y-a)/sqrt(b), so dy/dx = sqrt(b) \n=> dx = (1/sqrt(b)) dy (this isn't strictly correct but it works here)\n\nso:\nI = int(exp(-(x-a)^2/b)dx,-inf..+inf) \nI = int(exp(-((x-a)/sqrt(b))^2)dx,-inf..+inf) \nI = int(exp(-y^2)*(1/sqrt(b))dy,-inf..+inf)\nI=(1/sqrt(b))*int(exp(-y^2)dy,-inf..+inf)\nI=(1/sqrt(b))*sqrt(pi)  (from our given result)\nI = sqrt(pi/b)\n\nThe seond part requries a decent understanding of integration by parts, and if you email me I can go through this with you.
424492
Redo the integral.  \nFor #1 I got pi*(3pi-4)/2\nint(2pi*sin(x)((3/4)pi-x))\n= 2pi*int((3/4)pi*sin(x) - x*sin(x)\n= 2pi*[-(3/4)pi*cos(x) - sin(x) + x*cos(x)]\nFor pi/2:\n2pi[-(3/4)pi*0 - 1 + (pi/2)*0]\n= -2pi\nFor 0:\n2pi*[-(3/4)pi*1 - 0 + 0]\n= 2pi(-3/4)pi\n= (-3/2)pi^2\nNow\n-2pi - (-3/2)pi^2\n= -2pi + (3/2)pi^2\nYou can simplify this to the above answer by factoring out pi and then getting a common denominator of 2.  (It probably isn't necessary though)\n-2pi + (3/2)pi^2\n= pi*(3pi-4)/2\n\nFor 2, I got pi*(pi-2).\nThe integration will be similar to the one above, so I won't type it out again.  You can message me if you have a problem with the second integration.
83265
That's an elliptic integral.  Check out the link.
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
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
957759
a. 0 = 7-x^2, x^2=7, x = +/-sqrt(7)\nArea = Integral[-sqrt(7), sqrt(7)] (7-x^2)dx = 7x - (1/3)x^3 |-sqrt(7), sqrt(7) = 2[7sqrt(7) - (1/3)sqrt(7)^3] ~= 24.69 u^2\n\nb. 0 = x^4-8, x^4 = 8, x = +/-8^(1/4)\nArea = (1/5)x^5-8x |-8^(1/4), 8^(1/4) = 2[(1/5)8^(5/4)-8(8)^(1/4)] ~absv= 21.53 u^2
400264
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
1106006
Let u = cx+d, then du = c dx and so (1/c) du = dx\nSolve for x\nu = cx + d\nu - d = cx\nx = (u-d)/c\nThe new integral is\n[a*((u-d)/c) + b]/u (1/c) du\n= (1/c)[(a/c)u - ad/c +b]/u du\n= (1/c)[(a/c) - ((ad/c) + b)(1/u)] du\n= (1/c)[(a/c)u - ((ad/c) + b)ln(u)] + K, where K is a constant\nReplace u with cx+d for your final answer.
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
700700
try putting x^2 = i*t\nwhere i = complex iota\n t is varible \n\nthere integral will be in form : i/2*( cos(t)+i*sin(t) ) dt\n\nso answer will be : 1/2 * (i* sin (t) + i*cos(t) ) = 1/2 * e^ (i*t) = 1/2 * e^x^2\n\nThat's the answer....... :D\n\n\n[Assuming u know that e^(i*t) = cos(t)+i*sin(t) ]\n\ncan be solved easily ......... try it
1060628
simple find the integral of pi*(e^(-x^2))^2 from a=0 and b=2 since as x goes to infiniety is goes to zero but it is very close to zero when x=2 you can find this by graphing f(x) \nanswer should be 0.9188 good luck!
1092533
The previous person almost got there.\nThe indefinite integral is (1+x)[log(1+x) - 1] + C\n\nSubstitute x=1 to get 2[log(2) - 1] + C\nSubstitute x=0 to get  -1 + C.\n\nThe definite integral is \n2[log(2) - 1]  - [-1]\n= 2log(2) -1\n=log(4) - 1\n=log(4) - log(e)\n=log(4/e)
1156022
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out. \nIf you get stuck, send me an e-mail or add additional details.
1122457
1st question:\ny = x+10\ndy = dx\n\nint((y-6)e^(y-10)/y^7)dy = (e^-10)int(e^y/y^6)dy - 6(e^-10)int(e^y/y^7)dy\n\nTo the side (for integration by parts for the first integral on the right side):\nu = 1/y^6, du = -6/y^7dy\ndv = e^ydy, v = e^y\n\n= e^(-10)((e^y/y^6) - int(e^y)(-6/y^7)dy) - 6(e^-10)int(e^y)/(y^7)dy +c\n= e^-10(e^y/y^6) + 6e^-10int(e^y/y^7)dy - 6e^-10int(e^y/y^7)dy + c\n= e^(y-10)/y^6 + c\n= e^x/(x+10) + c\n\nSecond question:\nDifferential on both sides\n\nyx^(y-1)dx + x^y/ln(y)dy + xy^(x-1)dy + y^x/ln(x)dx = 0\ndx(yx^(y-1) + y^x/ln(x)) + dy(x^y/ln(y) + xy^(x-1)dy) = 0\ndy(x^y/ln(y) + xy^(x-1)) = -dx(yx^(y-1) + y^x/ln(x))\ndy/dx = -(yx^(y-1) + y^x/ln(x))/(x^y/ln(y) + xy^(x-1))\n\nYou really ought to ask these as separate questions though.
883154
I assume that's suppose to be an integral symbol.\n\nf(x) = [1/2 t²] from 2 to x, or 1/2 x² - 1\n\nThen f'(x) = x, and f''(x) = 1.\n\nDidn't really need to do the integration...
94569
Your integral needs either limits (from 1 to X, for example) or a constant added to it. With limits, the apparent contradiction vanishes. With an arbitrary constant, it can be accounted for.
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C