i had numerical approximation to a function say (x), and I want to integrate that function
217277
If you have a numerical approximation, your integration can not be more accurate than approximate, right?  And that may be close enough.  Integration is often the process of finding the area (integrating) under a curve, although the "curve" may be linear.  If you can draw your curve on a graph where for every value of x there is a single value of y, you can divide the graph into vertical slices where the bottom of each slice is the x axis and the top of each slice is a short horizontal line at the "eyeball" average height of the y axis in the slice.  If you divide the x axis into say five slices you then calculate the area in the five slices (A1 = x1 x y1, A2 = etc.).  You can increase the accuracy by using say 10 or 20 slices.  Of course calculus uses an infinite number of slices rather than approximation.\n\nAs a simple example, you can plot a "curve" of velocity vs time, where the velocity is Y and time is X.   In case one, velocity is constant (60 mph) and time increases to say 5 hours.  Y is a straight horizontal line at Y = 60 MPH from X = 0 to X = 5 Hr.  The "area" under the curve is a rectangle; 60 MPH x 5 Hr = 300 Miles.  So the integration represents distance traveled.\n\nIf in case two velocity increased linearly from Y = 0 to Y = 60 as time increased from X = 0 to X = 5, the area under the curve would be a triangle and can be calculated as 1/2 base time height or 1/2 (5 - 0) x (60 - 0) = 150 (miles).\n\nWhen the function is a real curve, you still need to estimate the area under the curve (with slices?) to integrate it.  You could have used slices in the above two examples to get an approximate answer but calculating area of a rectangle or triangle was easier and more accurate.  Good luck.
66930
Suppose that the limits on your integration are from 1 to X\n\nint{1/x dx from 1 to X} = (1-1)+int{1/x dx  from 1 to X}\n\nand one finds out that 0=0\n\nIf you want to perform an indefinite integral,\n\nint{udv}=uv-int(vdu} +C, where C is a constant.\n\nif u=1/x, and v=x, then choosing C=-1 makes this equality true.
277549
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
29162
http://integrals.wolfram.com/
353768
This *is* the way to write it so that x is the input and w is the output. If you want to solve for x in terms of w (i.e. have w the input and x the output), you are better off doing numerical approximations than using the cubic formula (it is really, really nasty).
561823
IF you want to try and get it in that EXACT form dicatated by Mathematica, I suggest you start with u=1+sin(x) and then solve for x in terms of u and do some resubstitution.  You'll get the arcsin function, which becomes a polynomial in terms of another dummy variable when you start the first u*dv substitution.  Notice I said "FIRST", because to get it in the form that you want, you'll have to do it about 6-10 times.  Also, you'll have to do some "right triangle simplification" throughout the problem as well.  Of course, then you get to factor it.  Have at it.  I did this very thing, stopping at the 3rd u*dv substitution because it is just ridiculous.  \nI did, however, get several of the terms.  I think if you decide to do this, you're going to need lots of paper.  I was doing it symbollically in a CAD program to keep things organized and such, and it was already 5 pages.  Oh, and not to mention that this "answer" given by Mathematica isn't even a closed form "answer".  It left it in terms of a polylogarithm, which is still an integral hidden in notation.  \n\nThe point is that the fuction (among others) is best integrated numerically because there is no useful closed form for this function.  This is what I did and it took me 2 minutes.  \n\nx^2/(1+sin(x))=x^2 - x^3 + x^4 - (5/6)x^5 + (2/3)x^6 - (61/120)x^7 + (17/45)x^8 - (277/1008)x^9 .................\n\nIntegrate termwise, which is fricking easy.  \n\n(1/3)x^3 - (1/4)x^4 + (1/5)x^5 - (5/36)x^6 + (2/21)x^7......\n\nNotice it is not undefined (like the original function) and the error term goes to 10^-7 after the first 5 terms.  I'd say that's as accurate as anyone would ever really need to be, and if not, they can always expand this function further.\n\nTell your uncle he's a bastard just for me.  He's teasing you.
452986
Let u = -x^2.\n\ndu = -2xdx.\n\n-1/2 du = xdx\n\nSo Integral [x*e^(-x^2)]dx = Integral[-1/2 e^u]du\n\nWith the substitution done, I think you should be able to evaluate the integral and back-substitute.
596795
Think about your order of operations. If you were evaluting the function for some number, what would you do first? That function is your *inside* function.
440788
Take the integral of this function, like this:\n\n∫ (10 + x²) / x³ =\n\n∫ (10 / x³) + ∫ (x² / x³) =\n\n10 ∫ (1 / x³) + ∫ (1 / x) =\n\n10 (-1 / 2x²) + ln(x) + C = \n\n-5 / x² + ln(x) + C\n\n~ ♥ ~
143990
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)
545751
"Main Idea: If f(x) is a function that has a derivative at a point x = a (i.e., f'(a) exists), then the line tangent to the graph of f at x = a, given by:\n\ny - f(a) = f'(a)(x - a);\nor, equivalently y = f(a)+f'(a)(x - a) can be an exceptionally good approximation of f(x) as long as x is close to a.\n\nAs a definition, we call the tangent line\nL(x) = f(a)+f'(a)(x - a)\nthe linearization (or linear approximation) of f at a. The\npoint x = a is the center of the approximation." (Ref 1)\n\nSo for f(x) = lnx at x = 8\n\nL(x) = ln8 + 1/8 (x - 8) (as f(8) = ln 8 and f'(x) = 1/x so f'(8) = 1/8)\n= 1/8 x + ln 8 - 1\n\nThis is useful to say estimate ln 8.1 if you know ln 8 (≈ 2.07944)\n\nSo ln (8.1) ≈ (8.1)/8 + 2.07944 - 1\n= 1.0125 + 2.07944 - 1\n= 2.09194\n\nBy calculator ln 8.1 = 2.09186..... so its pretty close (8^O)
113568
i made a little comment for you on your integral answer:)
271178
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
497666
Take the derivative of the original function. For your given example, if I remember my calculus correctly:\nf(x)=y\nf'(x)=1\nGraph the new equation.
532321
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
89076
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!
352637
dy/dx=2xy\ndy/y=2xdx\nintegrate\nlny=x^2+C\nput in initial conditions\nln(1)=1+C\nC=0-1\n'e' each side\ny=e^(x^2-1)\n\ncheck in original equation\ndy/dx=2x*e^(x^2-1)\n2x*y=2x*e^(x^2-1)\nthis answer is correct\nat x=1.6\ny=e^(1.6^2-1)=4.758821245\n\nto apply euler's method\nto the initial value problem\ndy/dx=f(x,y),  y(x0)=y0\nproceed as follows\n(a)take x0 and Y0=y0 as starting\nvalues,choose a step size h,\nand set r=0\n(b)set x(r+1)=r+h\n(c)calculate an approximation \nY(r+1) to(y(r+1)),using the\niteration formula \nY(r+1)=Yr+hf(xr,Yr)\n(d)if furtherapproximate values\nare required,increase r by 1,and\nreturn to sub-step (b)\n\ni also have details of the \nclassical forth order runge-\nkutta method,but will keep \nthat for a later date\n\ni hope that this helps