what is the integral of 1/(x^2)?
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
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
561879
The series for (1-cos(x))/x^2 is \nsum of (-1)^2n * x^(2n-2) divided by (2n)! (using Taylor series and simple algebra) which is:\n\n(1/2)-(x^2/24)+(x^4/720)-(x^6/40320)+.....\n\nNotice that this series is continuous on the open interval (0,1) which is good for integration.\n\nNow integrate to:\n\n(x/2)-(x^3/72)+(x^5/3600)-(x^7/282240)+......\n\nevaluate at x=1 and x=0 (however notice that the x=0 doesn't commit anything to the evaluation, so only evaluate at x=1)\n\nYou get 1/2-1/72+1/3600-1/282240+...\nNotice that each term contributes less and less to your answer.  Find the term that is less than 10^-5.  \n1/2=0.5\n1/72=0.014\n1/3600=2.778*10^-4 (getting close)\n1/282240=3.543*10^-6 (here it is)\nSince the term above only contributes to the ten-millionths decimal place, you can safely assume you have found an estimate of the original function using a series out to the hundred-thousandths decimal place, or have an error of less than 10^-5.  \n\nEnjoy!
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!
667880
a) dy/dx = x - 1/x^2\ndy = (x-x^(-2))dx\ny = x^2/2 + 1/x + C\ny(1) = 1^2/2 + 1/1 + C = 2\n1/2 + 1 + C = 2 implies C = 1/2\nAnswer: y = x^2/2 + 1/x + 1/2, x>0\n\nb) The integration is already finished.  Just plug in -1 for x.\ny(-1) = (-1)^2/2 + 1/-1 + C = 1\n-1/2 - 1 + C = 1 implies C = 3/2\nAnswer: x^2/2 + 1/x + 3/2, x<0\n\nc) Just take the derivative of each one.  C1 and C2 are constants, so their derivative is 0.  \n\nd) C1 = 1/2 and C2 = 3/2\n\ne) Follow parts (a) and (b) except plug in 2 for x and set it equal to -1.  Similarly for y(-2) = 2.\nCheck that C1 = -7/2 and C2 = 1/2
727341
first integrate\n\nF(x) = x^3 -x^2  from [1, 3]\n= 3^3 -3^2 -1^3 + 1^2\n= 27 - 9 -1 + 1\n= 18\n\naverage = 18/(3-1) \n= 18/2\n= 9
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
6015
intgral(from - infinity to infinity) of e^(-x^2)= sqrt(pi)\n\nThis is the area under the bell curve. It comes up very often in statistics.
916838
(1/2)(1/2)=1/4\n\nB. 1/4
677393
it is probably easiest to expand into a polynomial like bboy does. or use u substitution:\nlet u =1-3x\ndu=-3 dx\nso integral of -2/3 u^2 = -2/9u^3 = -2/9(1-3x)^3
822637
When I have problems with an indefinite integral I use some of my symbolic software to get straight to answer or a "Table of Integrals".  And then I focus on how to get to that particular answer, whether it is through substitution, changing it to a double integral in polar coordinates, etc.\n\nA "free" online integrator program that is offered by Wolfram (maker of Mathematica) is located at\n\nhttp://integrals.wolfram.com/\n\nType your particular problem as\n \nArcTan[1/x] \n\nand it will give you an answer that is solved symbolically for an "indefinite" integral which is\n\nx(arctan(1/x)) + (1/2)(log((x^2) + 1))\n\nHowever to be honest with you it was a few years back when I saw this particular integral, and I honestly don't remember the process.  I think it was a definite integral though.
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
246551
integral of sine squared.\nso theres a conversion thing which says \nsin^2 x = (1-cos 2x)/2\nso if you use that then u have to integrate 1/2 - cos 2x/2\nwhich is just 1/2x - sin 2x / 4
147855
Try http://integrals.wolfram.com/
821024
∫cos(x^2) dx = √(π/2)FresnelC(√(2/π)x)
885147
You have to note that if f(x) = arctan(x), then f ' = 1/(1+x^2).\n\nFurthermore integral (2 f '  f dx) = f^2.\n\nTherefore integral (arctan (x) /(1+x^2)) = (1/2) (arctan(x))^2
1090386
What?\nIf u=x^x\ndu/u=(1+ln(x))dx\n\nInteral  exp(x^x)dx=integral exp(u)du\n\nAnalytical solution for this indefinite integral is dificult.