what is the definition of an integral?
919724
Two things, which turn out to be identical:\n - The area under a curve.\n - The anti-derivative of a function, i.e. a function whose derivative is the given function.
219346
The first link below is a problem discussing improper integrals (which classifies all integrals with an infinite limit of integration). Just scroll to the bottom of the page--the top of the page is a different problem altogether. Basically, the tip is to replace the "troublesome" limit values with a constant (like k) and then evaluate the limit as k approaches that value. In this case, k becomes one of the limits of integration and you evaluate the integral as k approaches infinity.\n\nI recommend The Complete Idiot's Guide to Calculus (follow the second link in the Sources list below) when brushing off calculus rust, and not only because I am the author of the book. :) It's a light-hearted but content-heavy review of calculus, beginning with advanced algebra review and ending with function approximation by infinite series approximation.\n\nAlso, spend some time with "Tutorials for the Calculus Phobe," free Flash-based calculus tutorials on my website, Calculus-Help.com. They're fun but help you conceptualize calculus basics. The third and final link in the sources houses the Phobe tutorials.\n\nBest of luck!
414817
int = integral\nint[(sinx)^5dx]\nRewrite (sinx)^5 as sinx (sinx)^4\n(sinx)^4 = ((sinx)^2)^2\n(sinx)^2 = 1 - (cosx)^2\nThus, \n(sinx)^4 = ((sinx)^2)^2\n= (1-(cosx)^2)^2\nNow the integral is\nint[sinx(1-(cosx)^2)dx]\nUse substitution.\nLet u = cosx, then du = -sinxdx\nNow the integral is\nint[-(1-u^2)^2du] \nExpand (1-u^2)^2 and multiply through by the -1\n-(1-u^2)^2 = -(1 - 2u^2 + u^4) = -1 + 2u^2 - u^4 \nThe integral now equals\nint[(-1 + 2u^2 - u^4 )du]\n= -u +(2/3)u^3 - (1/5)u^5 + C, where C is a constant\nReplace u with cosx to get the final answer.\n-(cosx) +(2/3)(cosx)^3 - (1/5)(cosx)^5 + C
274258
An integral with which the limits are known, as oppossed to an indefinite integral.
47759
sadly, there is no easy way to remember how to do integrals.  integrating is a major, bigtime pain in the ass.  for this reason people buy books of tables of integrals - they are quite literally a whole book filled with integrals of various forms, with solutions.  very useful if you have to do them a lot.\n\nalso, wolfram research, who create the mathematica software, have an online integrator that usually works (but sometimes gives you ridiculous answers, so use with care):  http://integrals.wolfram.com/\n\nas for programming - numerical methods for doing calculus are pretty easy.  if you have a function and want to integrate it, just divide it up into a bunch of rectangular slices, where the height of the rectangle is the value of the function in the middle of the slice, and the width of the slice is as small as you need it to be.  then just add all the areas up.  the smaller the width, the more accurate the calculation will be, but the longer it will take, so your accuracy is limited only by your patience and the speed of your computer.\n\na slightly more accurate method is to approximate the top of the rectangle with a parabola - this is called simpson's method:  http://spiff.rit.edu/classes/phys317/lectures/num_integ2/num_integ2.html\n\nthough quite honestly, computers are so fast these days that it's hardly worth the bother of figuring it out.  just using more rectangular slices and letting it run for a few minutes should almost always do the trick.\n\ncheers
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
245944
The integral of a function over a definite interval.
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.
586082
integrate by parts.\n\nintegral(lnx dx)\n= (lnx)(x) - integral (x d(lnx))\n= xlnx - integral (x (1/x) dx)\n= x lnx - integral (1 dx)\n= x lnx - x + c
274471
i made a little comment for you on your integral answer:)
318235
Integration is the calculation of the area under a curve and the inverse of integration - differentiation is the calcutation of the gradient of a curve
894395
∞\n\nConsider that this function is everywhere positive, and that as the lower bound of the integral increases, the function being integrated does not approach zero (in fact, it becomes increasingly large), so the integral of e^(y²) from -∞ to anything must in fact be infinite.
517722
There is more information about "Cauchy Integral Theorem" at this site: http://en.wikipedia.org/wiki/Cauchy_integral_theorem\n\nCauchy integral theorem\n\n\nIn mathematics, the Cauchy integral theorem in complex analysis, named after Augustin Louis Cauchy, is an important statement about path integrals for holomorphic functions in the complex plane. Essentially, it says that if two different paths connect the same two points, and a function is holomorphic everywhere "in between" the two paths, then the two path integrals of the function will be the same.\n( http://en.wikipedia.org/wiki/Cauchy_integral_theorem )
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
859250
integral = x*tan(x)+ln(sec(x))  (use integration by parts with du = sec^2(x) and v = x).\n\ng'(x) = f(x), so a, c, e and g are true (b is true if f is continuous)\n\nSteve
833486
int1/x^2+int1/x^5\nx^-1/-1 +x^-4/-4+C\n-1/x-1/4x^4+C
737583
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards