How do you can integrate a function that is represented in numerical approximation?
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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.
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An integral with which the limits are known, as oppossed to an indefinite integral.
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There are a number of available tools in the following website:\n\nhttp://www.functionx.com/
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See this site for details of lagrange polynomials, then the program shouldn't be hard to write:\n\nhttp://en.wikipedia.org/wiki/Lagrange_interpolation
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Integrate from -π/2 to π/2 and double it
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1. um, everywhere? From a cash machine, to calculating how much chemicals to put to produce laundry detergent, to construction of buildings and bridges.\n2. The ever-increasing advances in computer technology has enabled many in science and engineering to apply numerical methods to simulate physical phenomena. Numerical methods are often divided into elementary ones such as finding the root of an equation, integrating a function or solving a linear system of equations to intensive ones like the finite element method. Intensive methods are often needed for the solution of practical problems and they often require the systematic application of a range of elementary methods, often thousands or millions of times over. In the development of numerical methods, simplifications need to be made to progress towards a solution: for example general functions may need to be approximated by polynomials and computers cannot generally represent numbers exactly anyway. As a result, numerical methods do not usually give the exact answer to a given problem, or they can only tend towards a solution getting closer and closer with each iteration. Numerical methods are generally only useful when they are implemented on computer using a computer programming language . \n\n3. Visit these sites:\nhttp://math.fullerton.edu/mathews/n2003/FixedPointMod.html\nhttp://en.wikipedia.org/wiki/Linear_interpolation\nhttp://mathworld.wolfram.com/NewtonsMethod.html\n\nHope that helps!!
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This integral does exist (it can be evaluated for any endpoints you choose). Since it cannot be represented as a finite combination of elementary functions and operations, it is given its own name, the error function. Many integrals like this have their own names and are quite useful in applied mathematics. See http://mathworld.wolfram.com/Erf.html .\nTo say that it does not exist is unrigorous.
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Homework.\n\nShow it first for intervals. Then extend to simple functions. Then use the fact thats imple functions are dense in L1.
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The integral of a function over a definite interval.
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In the mathematical subfield of numerical analysis, interpolation is a method of constructing new data points from a discrete set of known data points.\n\nIn engineering and science one often has a number of data points, as obtained by sampling or some experiment, and tries to construct a function which closely fits those data points. This is called curve fitting. Interpolation is a specific case of curve fitting, in which the function must go exactly through the data points.\n\nA different problem which is closely related to interpolation is the approximation of a complicated function by a simple function. Suppose we know the function but it is too complex to evaluate efficiently. Then we could pick a few known data points from the complicated function, creating a lookup table, and try to interpolate those data points to construct a simpler function. Of course when using the simple function to calculate new data points we usually do not receive the same result as when using the original function, but depending on the problem domain and the interpolation method used the gain in simplicity might offset the error.\n\nIt should be mentioned that there is another very different kind of interpolation in mathematics, namely the "interpolation of operators". The classical results about interpolation of operators are the Riesz-Thorin theorem and the Marcinkiewicz theorem. There also are many other subsequent results.
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Which function, Subsitution, Linear COmbination, Graph?
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There is no function that exists that has as its derivative e^(x^2).  As someone else pointed out, you can use the Maclaurin series representation and integrate term by term.  However, this series converges very slowly and would require a lot of terms for two place decimal accuracy, especially if x is large.
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Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.
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try the link below...... hope it can put an end to ur integral problems
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Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
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the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated
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http://mathforum.org/library/drmath/sets/high_calculus.html