What is a Contour Integral?
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In general, Integration is a summation process of [ f(θ) dθ ], where f(θ) is a function and (d θ) is the element of the path (or axis)  that the integration is carry over.\n\nLimt ∆θ →0  ∑n=∞ n=1  f(θ) ∆ θ = ∫ab  f(θ) d θ …. \n\nWhere ∆ θ = (b-a) / n\n\nNow, the path maybe (x-axis) then f(θ) should be function of x only and d θ became dx in this case the integral called integration over x-axis.\n\nor the path maybe (y-axis) then f(θ) should be function of y only and d θ become dy and the integration called integration over y-axis.….. and so on for any other path.\n\nThen for a contour (L) , which is a line that is not parallel to the coordinates axises, the f(θ ) should be function of L and d θ is dL; and in this case the integration is called Contour Integral.\n\nUsually the contour integral is used to fine the work done due to the moving of an object by a force. Also, the contour integral can be use to find the length of a line.
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Let S be a surface in three-dimensional space ( R^3 ) and let f be a function defined on this surface. To describe the surface, we will choose a parameterization of the surface by two variables which we shall call u and v. (For instance, if f is a sphere, u and v might be the latitude and longitude.) In terms of this parameterization, the function f can be expressed as a function of u and v. Then the integral of f with respect to the surface area is usually notated as \n\nSurface Integral S[f(u,v)d^2 A]\n\nJust as with the ordinary integral, this quantity may be understood as a limit of sums.1 To form an approximating sum, we subdivide the surface S into a miniscule pieces, multiply the area of each piece by the value of f at a point located inside that piece, sum over all the miniscule pieces into which we have subdivided the surface. The limiting value of these sums as the size of the pieces shrinks to zero is the integral with respect to surface area. \nTo compute this quantity, one may use the following formula to convert it to a double integral: \n\n\nIf the surface is described as the graph of a function , then we may also use the following formula:
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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
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This integral does not have an indefinite closed form of elementary functions. It can be evaluated numerically by either integrating the series representation of e^(-x^2) or using integral approximation methods like Simpson's rule.\nIt also has an interesting value if evaluated over its entire domain; by changing coordinates to polar coordinates, one can easily show that the integral over the entire real line is equivalent to Sqrt[Pi].\nThis fact is used in statistics to normalize the Gaussian distribution to have area 1 under the curve. The integral of your expression on the interval [0,t] is then known as the error function. See the source for details.
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By integrating the function between the limits you want the area under.
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This is a non-elementary integral in x. However, from the way you phrased the problem, I would guess that you want this to compute a double integral: first with repspect to x, then with respect to y. The best way of doing *that* would be to switch the order of integration. Then the integral with respect to y will be xe^(y/x). Be sure to change the limits of integration so that you are integrating over the same region in the xy-plane!
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The solution of an ordinary differenatial equation is the same as finding the family of curves whose tangent vectors agree with the vector fields generated by the ordinary differential equation. These curves are called integral curves, and the solutions are called integrals of the differential equation.\nThe analogy is primitive. In multivariable calculus, you may have integrated terms of the type f(x, y) dy dx (an area/surface integral) or f(x) dx + g(y) dy (a line integral). We unify the study of all these types of integrals through the study of general differential forms (objects which look like sums and products of the terms mentioned above) and their integrals. This study is equivalent to the study of partial and ordinary differential equations.\nAs an example, the form f(x) dx + g(y) dy is saying that the tangent vectors to the solution curves have an x-component that is f(x) and a y-component that is g(y). \nAs a specific example, the equation of a unit circle is x^2 + y^2 = 1. The differential of this equation is 2x dx + 2y dy = 0. This is an ordinary differential equation (you can algebraically rearrange the differentials of the coordinate functions dx and dy to get a standard form if you want). \nSimilarly, we can have some physical problem that inspires us to examine the equation 2x dx + 2y dy = 0. It is common to know dynamic behavior and have to figure out the static relationships between the variables involved. The tangent field is made up of the vectors (2x, 2y) attached to each point (x, y). We integrate both sides of the equation to get x^2 + y^2 = C where C is an undetermined constant. Thus, we call the curves that satisfy this equation integral curves of the vector field.\nIf it is not easy to find an analytic solution to the vector field, we may use a numerical approximation. Since we are approximating the integral solution, we call it numerical integration.
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If your curve is the upper half of an ellipse, then when you rotate it, you have half an ellipsoid.\n\nWikipedia has the formula for the surface are of an ellipsoid, which (surprize!) involve eliptic integrals.
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"Contours, which are lines drawn through points the same height above sea-level, allow both the direction and gradient (steepness, see page 18) of slopes to be determined from the map with much greater accuracy than either of the above techniques" [Hachures is the only other technique I can see.] "Contours are included on almost all modern topographic maps.  The vertical distance between two adjacent contour-lines is known as the vertical interval (V.I.).  A V.I. of 20 metres is often used on modern metric maps.  On older pre-metric editions a V.I. of 50 feet was widely used.\n"When contours are close together, slopes are steep, merging when the gradient (steepness, see page 18) approaches verticality, whereas on gentle slopes the contour-lines are well spaced (Figure 5).  Where slopes are even, the contours on a map are evenly spaced, while on irregular slopes the distance between contours varies widely."\nMap Reading by Patrick Armstrong, Publishied in Australia by Rigby Limited 1980, p15.\nISBN  0 7270 0999 0\n\nI cannot reproduce Figure 5, nor will I explain gradient from p18.  Note that it is an OLD book, published 26 years ago, so when it says modern, it does not mean modern today.
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integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
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put x^2=t\n2xdx=dt\nxdx=1/2dt\nthe integral now is\n(1/2dt/1+t^2\n(1/2)tan^-1t+C\n=(1/2)tan^-1(x^2)+C
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Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
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http://mathworld.wolfram.com/PoincareConjecture.html
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I have never used them, but I have heard they have uses in certain kinds of electrical engineering.  I think it involves phase shifts of period functions like sine waves.  Also, there are occasional problems where you start in the real numbers and end in the real numbers, but by doing the problem in the complex domain, certain difficult problems become solvable.  I think Cauchy integrals are used for this purpose.
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My old Calculus teacher had this pneumonic that was a good guide:\nL.I.P.E.T.\n\nFollow in the order of the letters to assign to u.  Look for first one in this list that your integral contains. Then assign that one to u:\nL-logarithmic functions\nI-inverse trigonometric funtions\nP-polynomial functions\nE-exponential functions\nT-trigonometric functions\n\nThen, assign everything that's left to dv.\nNext, take the derivative of u in respect to x, but move the dx over to the other side with x.\nThen, find v by integrating dv. For our purposes in finding v, leave off the constant C that you'd usually get for an indefinite integral.\nNext, use the following formula:\nIntegral of:(u dv) = uv - Integral of:(v du)\n-Where the Integral of:(u dv) is just your original integral.\n\nRepeat this process for the next integral if necessary.\n\nExample: \n("^" means raised to the power, e is the exponential constant)\n\nIntegral of:(xe^x dx)\nLet: u=x (x is a polynomial, so assign this to u; e^x is lower down on the list, so we don't use that for u)\ndu=dx\nLet: dv=e^x dx\nv=e^x\nIntegral of:(xe^x dx) = xe^x - Integral of:(e^x dx)\nIntegral of:(xe^x dx) = xe^x - e^x + C\n(Don't forget your constant C at the end)
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Calculus was confusing enough in English - I can't imagine taking it in Japanese.  You might want to try to get hold of a textbook or something in English.\n\nIt's an integral.  More specifically an indefinite integral.  It's kind of like the inverse of the derivative function except for that there is no way to get the constant back (hence the "C").  Basically the indefinite integral is the set of all functions whose derivative is equal to the thing between the integral sign and the dx (or dy, dq, whatever the variable is... I'm blanking on general name of the thing).  The integral symbol tells you it's an integral (obviously), and the "x" in dx tells you what variable you're integrating with respect to (that will make more sense when you learn to integrate more complicated expressions).  So the indefinite integral of x^2 is 1/3x^3 + C with respect to x where C is some constant because the derivative of 1/3x^3 + C is x^2 (remember that the derivative of a constant is zero).\n\n***A Note on Definite and Indefinite Integrals***\nA couple of the answers below appear to be describing the definite integral.  The definite integral can be described as the sum of the areas of rectangles, and can be used to find the area under a curve (among other things).  The indefinite integral (or antiderivative), however, which is the one refered to in the question, describes a function.  Although the function described by the indefinite integral is extremely useful in calculating the area under a curve, it does not directly describe this area.  In addition to being used to caculate the definite integral, the indifinite integral can also be used, for example, to find an equation for the position of an object given an equation for its velocity.\n\nAnother afterthough: I may have gotten some of the terminology wrong as my high school math department had a strange affinity for inventing their own terms.
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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...