Kidnapped! WebQuest

Introduction to Riemann Sums

One fundamental problem that led to the development of Calculus was that of finding the area under an arbitrary curve. George Bernhard Riemann's idea was to use thin rectangles as illustrated to approximate the area.

> with(student):

> leftbox(sin(x)+x,x=0..10,20);

In the above example, we have chosen to use the left endpoint of each subinterval (rectangle) as the point which should touch the curve. We also could have chosen the right endpoint, the midpoint, or even a random point on the top edge of each rectangle.

> rightbox(sin(x)+x,x=0..10,20);

> middlebox(sin(x)+x,x=0..10,20);

From the illustrations, we can conclude that the left Riemann sum will underestimate the area under the curve, the right Riemann sum will overestimate the area under the curve, and the midpoint Riemann sum will be the most accurate. This is not always the case. Depending on the individual characteristics of each curve, any of the methods could be the most accurate. Keep in mind, if we were to take the Riemann sums with more subintervals (we have only used 20 in our examples), we would get a better answer in any case. In fact, as the number of subintervals approaches infinity, each method will converge to the exact area.

It is this exact area under a curve that is represented by the definite integral

> Int(sin(x)+x,x=0..10);

But before we learn how to directly evaluate the above expression, we will use the Riemann sum and other techniques to approximate the definite integral (the area under the curve). Later we will investigate the physical meaning of this area.