The solutions to Practice Exam #4 are now posted.

Don't let this happen to you this week!

Problem 9 in Section 6.1 is an example of an area for which it is much easier to compute the area by integrating in the y-direction than in the x-direction.

If you click here, you can see the problem done out the "easy way" (integrating dy) and the "cowboy way" (integrating dx).

The technical difficulties plaguing this site have been fixed. Now I know what happens when you accidentally save files to the wrong directory.

Please let me know if you have any troubles getting to materials, and I apologize for inconveniencing you.

Lots of stuff is happening in MA121 during this last full week of class.

• The solutions to Quiz #9 have been posted to the Exam and Quiz Solutions page..
• Exam #4 happens this Friday, December 5th. It covers Section 4.11, all that we've done in Chapter 5, plus the area and volume stuff from Chapter 6.
• Here are some practice problems for the exam.
• One final project assignment, which will count for a lab grade. It is due on Tuesday, December 9th in class. It is a team assignment, either two or three people per team, no crossovers between sections.

I will talk about the final exam next week, after we get over this hurdle.

Not far to go now! Only 1-1/2 weeks left in the term, and we have one project, one test, and one final exam.

The most recent quiz was returned. I hoped it would go better than it did ... maybe it was just a product of lots of things due in other classes near the end of that week.

From now until the end of the semester, we'll discuss applications of integrals. Today's wasn't too hard. Our very first application was to find the area under a curve for a positive function. We can now expand to find the area of a region between two curves. By way of Riemann sums, we found that if f is the function for the top curve and g the function for the bottom curve, then the area of the region between them from a to b is given by



Sometimes, we're given all the information we need to set up this integral. Most times, we will have to discover the boundaries ourselves, as in the parabolas example we did in class. Usually, we just set the two functions equal to each other and solve for x to get our a and b values if they're not already given. One way to think of what we're doing is as if a laser scanner is being run across our region from left to right, measuring the length of its beam as we go... sometimes, as we saw, it's easier to run the scanner from bottom to top instead of from left to right.

Wednesday, we move on to computing volumes of particular types of solids using integration.

I hope everyone had a great Thanksgiving break and I hope you don't have to run as far as I do to counterbalance the amount you ate! In case you've forgotten, I am Professor Poodiack and this is the blog for your Calculus class.

Before the break in Calculus I, we finally finished up a long example started the previous Friday. (Actually, the class finished it while I enjoyed a hot steaming cup of coffee.) We then established some properties of definite integrals, some of which were just like properties of indefinite integrals.

Finally, a lot of hemming, hawing, and teasing, came to an end as we started talking about the Fundamental Theorem of Calculus, Part I. (Most textbooks refer to this as part I, even though MathWorld refers to it as the "second theorem." Maybe Wolfram ranks in terms of amount of use.) After slogging our way through antiderivatives -- indefinite integrals -- and Riemann sums for definite integrals, we're convinced that the similar names are not a coincidence.

The Fundamental Theorem of Calculus explicitly states the links between derivatives, antiderivatives, and definite integrals. In fact, Part I shows that essentially definite integrals and derivatives are inverse operations:

Fundamental Theorem of Calculus (Part I): Let f be a continuous function on [a,b]. Then the function



is continuous on [a,b] and differentiable on (a,b). Moreover,

Another way to say this is by way of Liebniz notation:


In other words, (kinda sorta) the derivative of the integral of f leaves us with f. In reality here, what we have is another derivative formula.

Before we broke, we discovered the ultimate link between derivatives, antiderivatives, and definite integrals by way of the fundamental theorem of calculus, part deux. After a couple of days of working to find limits of Riemann sums, we were ready for (and deserved) a shortcut.

Fundamental Theorem of Calculus (Part II)

Let f be a continuous function on the interval [a,b]. If F is any antiderivative of f, then



This is an incredibly deep result because it ties together a purely algebraic object (the antiderivative, or indefinite integral) with a geometric object (the definite integral, visualized as a sum or difference of rectangle areas).

We decided first off that since the function F can be any antiderivative, we'll always choose the one with constant of integration 0 (so we don't have to worry about the "+ C" part). Also we have some new notation:



So, for example, we can compute the area under the curve from x = 0 to x = 1 as



Our quest to reverse differentiation will now turn into finding ways of reversing well-known rules. The first one we're trying to reverse is the Chain Rule.

One way of handling the Chain Rule was by setting a new variable u to be the "inside" of our function -- a substitution, if you will. We will try the same trick. If we have a difficult integration to perform, we will try setting a new variable u to be the "inside" of the most difficult part. This substitution will hopefully transform our difficult integral into one of the ones that from our basic table of antiderivatives.

For example, to calculate , we set . This means that and



If we were, say, missing the factor of 3 in the differential, we'd have to compensate by inserting a factor of 1/3 when we switch the integral over.

For definite integrals, we have two ways to go. We can find the antiderivative as a side calculation. For example, we have



We can also change the limits of integration as we go and not have to worry about changing back to x's at the end. For the above example with , we know that if x = 0, then u = 1 and if x = 2, then u = 9. Therefore,



We'll start working on applications of integrals in Chapter 6. See you Monday.