The solutions to Quiz #8 are available from the Test and Quiz Solutions page.

Thursday in Calculus, we introduced an object called a definite integral of f from a to b as the limit of Riemann sums for f, where unlike the area formulation we used yesterday, f does not have to be nonnegative. That is,



where f is a continuous function on [a,b], the 's are sample points for the i th subinterval, and is the width of that subinterval. If f is nonnegative on [a,b], then the definite integral equals the area under the curve from a to b. If f becomes negative somewhere in that interval, the integral equals the area above the x-axis minus the area below the x-axis. If f is a velocity function, we saw today that the definite integral will give the exact distance traveled between time a and time b.

It is quite difficult to compute definite integrals as limits of Riemann sums. For example, suppose with a = 0, b = 3, and n = 6 right rectangles. Visually, here is what we have:

Pasted Graphic 5

We can see that there is more area below the x-axis than above, so we expect a negative answer. (We get - 3.9375 in fact.)

We figured out that for known areas (circles, rectangles, and triangles), we can find the value of certain definite integrals really easily, without having to deal with rectangles, sums, and limits. For the unknown areas, we have to go back to that technique.

On Monday, we'll find an easier way to compute the definite integral.

This week in MA121 we seemed to veer off the path we were on. Instead of continuing with antiderivatives, we talked about how to compute the area under a curve.

The idea is this: Suppose we have a curve between x = a and x = b, and that in that interval. We're going to try to compute the area of the region under our curve but above the x-axis. In most cases, we won't have a region whose area fits one of our well-known formulas (rectangle, triangle, circle). So in the typical calculus manner, we're going to sneak up on the answer.

We do this by overlaying the region with rectangles, the region whose area formula is the easiest we know. Formally what we do is split the interval from a to b into say, n, subintervals by picking partition points so that



(In practice we usually pick the points to be equal distances apart, but you don't have to. Sometimes in numerical integration it's advantageous not to do so.)

So we have these subintervals . Let's call the width of the first interval , the width of the second interval , etc.

In each of these subintervals, we pick one point called a sample point, where we're actually going to compute the value of the function. Let's name these sample points .

Now we draw rectangles with width and height . This means the area of each rectangle is . We can approximate the area under our curve by adding up the areas of all the rectangles we've drawn. That is,



This sum is what is called a Riemann sum. It will come up quite often in defining quantities for us down the line.

We also noticed that when we use more rectangles, our approximation is better. So we surmised that


as long as the limit exists. (Here and above, we have used sigma notation, which represents long sums of numbers.)

We will examine this sum in more detail on Friday. See you then.

The solutions to Exam #3 are now available from the Test and Quiz Solutions page.

I owe both sections an apology because other deadlines -- having nothing to do with teaching classes -- have put me way behind in terms of grading and blogging. (The derivative on papers graded has been 0 ... didn't move even one through the system last night.)

You won't get your exams back until Thursday, but I will post the solutions Wednesday morning. I will also try to catch up on blogging tomorrow. Your labs will definitely be returned before the end of the semester.

We will finish up 5.1 and do some work on 5.2 on Wednesday.