Search results
Results from the WOW.Com Content Network
Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. [1]
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx.
This can be proved by computing the derivative of the right-hand side of the formula, taking into account that the condition on g is here for insuring the continuity of the integral. This gives the following formulas (where a ≠ 0), which are valid over any interval where f is continuous (over larger intervals, the constant C must be replaced ...
The absolute value of the Jacobian determinant at p gives us the factor by which the function f expands or shrinks volumes near p; this is why it occurs in the general substitution rule. The Jacobian determinant is used when making a change of variables when evaluating a multiple integral of a function over a region within its domain. To ...
The method also is applicable to other multiple integrals. [1] [2] Sometimes, even though a full evaluation is difficult, or perhaps requires a numerical integration, a double integral can be reduced to a single integration, as illustrated next. Reduction to a single integration makes a numerical evaluation much easier and more efficient.
The Cauchy formula for repeated integration, named after Augustin-Louis Cauchy, allows one to compress n antiderivatives of a function into a single integral (cf. Cauchy's formula). For non-integer n it yields the definition of fractional integrals and (with n < 0) fractional derivatives .
A simple example of such a problem is to find the curve of shortest length connecting two points. If there are no constraints, the solution is a straight line between the points. However, if the curve is constrained to lie on a surface in space, then the solution is less obvious, and possibly many solutions may exist.
Naturally the analogues of contour integrals will be harder to handle; when n = 2 an integral surrounding a point should be over a three-dimensional manifold (since we are in four real dimensions), while iterating contour (line) integrals over two separate complex variables should come to a double integral over a two-dimensional surface.