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Name: Clarence
Status: student
Grade: 9-12
Location: CA
Country: USA
Date: Fall 2011

We know that the area under a continuous function on a certain interval, for instance, [a,b], can be found out by definite integration (taking limit of a sum of areas of rectangles). But when the curve is a discontinuous function, the limit does not exist and is not independent of the choice of the height of rectangle in each delta x. Does it mean that the height in an interval delta x including f(x) of undefined value can be any value so that a limiting value (area under the curve) does not exist?

I would not say that the height in the vicinity of the discontinuity can be ANY value, because you know the value of the function on either side of the discontinuity. It seems natural choose, as the height of the rectangle, the value of the function at the x value of the middle of the rectangle. In this case, the height will change as you decrease dx, depending on where the x midpoint of the discontinuity-straddling rectangle happens to lie.

In actual practice, of course, you rarely have to worry about this. The fact that there is a discontinuity usually means that you do not have a single functional form that represents the function on both sides of the discontinuity, so you just split the integral into two integrals and add them.

Tim Mooney

Even when the function is discontinuous there may be a well-defined integral. Take for instance a function f defined in [a,b] which has discontinuity at a point c, but this discontinuity is of the kind that the lateral limits exist. Then f is integratable and its integral equals the integral from a to c plus the integral from c to b. There is a theorem of Lebesgue that states that a bounded function f on [a,b] is (Riemann) integratable if and only if the set D of discontinuity points of f is of measure zero, in the sense that for any \epsilon>0 there exist denumerably many intervals (a_n,b_n) such that their union contain [a,b] and the sum of their lengths do not surpass \epsilon.

However, functions such as Dirichlet's (this function is defined on [0,1], equals 1 on rationals and 0 on irrationals) are not Riemann integratable, since its discontinuities are the whole [0,1]. If you wish to integrate these functions, if have to go for Lebesgue's theory of integration.

Conrado Lacerda

The integration of discontinuous functions is a large subject, too large to go into detail here, but do a Search on the term "integration of discontinuous functions". You will find the topic discussed on several sites at different levels of rigor, depending on how much detail you want to know. Some discontinuous functions can be integrated by dividing up the function into pieces so that it is "piece-wise" continuous, but this approach has its limitations. There are discontinuous functions that cannot be integrated.

Vince Calder

Clarence, The process of rectangles only applies when the limit exists. You must arrange your rectangles so that the limit is defined. This occurs when the discontinuity is held as a border between rectangles. Find the sum of the two limits, the two areas, one to the left of the discontinuity and one to the right. Sometimes, provided you choose you algebra wisely, you can find the integral even when a vertical asymptote exists within the range of the domain of your integral.

Dr. Ken Mellendorf Physics Instructor Illinois Central College

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