By Walter Thirring

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817–818 of Edwards’ book that I mentioned in the Preface. The result of 2π 15 is the answer derived by Edwards. 28 1 Introduction When we ‘check’ integrals that are just given to us, we’ll use quad or int, but when we are faced with this new type of question we need to do something different. What we’ll do still uses a computer, but now we’ll simulate the physical process of drawing a circle C2 using points chosen randomly from the interior of a given circle, C1. That is, our computer will ‘draw’ many such random C2 circles and literally count the fraction of them that are totally inside C1.

The Riemann integral is thus just a special case of the Lebesgue integral, as a sub-interval is just one particular way to define a set of points; but there are other ways, too. When the Riemann integral exists, so does the Lebesgue integral, but the converse is not true. When both integrals exist they are equal. To see how this works, let’s calculate the Lebesgue integral of ϕ(x) over the interval 0–1. In this interval, focus first on all the rational values of x that have the particular integer n in the denominator of the fraction mn (by definition, this is the form of a rational number).

But—one does have to be ever alert for pitfalls that, we hope, quad will be our last-ditch defense against. Here’s an example of what I’m getting at, in which that most common of operations—blindly plugging into a standard integration formula—can lead to disaster. ð1 À Á2 dx . The integrand f ðxÞ ¼ 1x is never Suppose we have the integral I ¼ 2 À1 x negative because it’s the square of a real value (from À 1 to 1). Thus, we know immediately, from the area interpretation of the Riemann integral, that I > 0.