Math 110.109, Calculus II (Physical Science and Engineering)

Spring 2011 Course Lecture Synopses

 

Week 1:  January 31  February 4, 2011

 

http://www.mathematics.jhu.edu/brown/courses/s11/109.htm

 

 

 

 

·       Monday, January 31:  Welcome to the course!  After a not-so-brief introduction to the purpose and focus of this course, as well as where it sits in your mathematical present and future, and how I will run it and such, I started on the content.  Starting in Chapter 7, Techniques of Integration, I used the idea that integration is the opposite of differentiation to establish the central focus of this chapter;  that integration, or anti-differentiation, in general is much more difficult, if even possible at times, to do.  But the patterns found in the differentiation rules we already have studied can lead to techniques of integration that are remarkably powerful in those cases where the pattern can be recognized.  Undoing the Power Rule (Section 3.1) , which works for all real numbers , is solely a matter of recognizing the pattern:  .  Sometimes this rule is referred to as the Anti-Power Rule, and works for almost all real numbers  (can you see where it doesn’t?  Do you know why it doesn’t?)  I started the discussion by bringing this up and then rehashing the Substitution Method, which I call the Anti-chain Rule.  The reason is that the pattern one looks for in an integrand to use this method, is ;  the integrand is a product of functions, one factor of which is a composition of two functions and the other factor is the derivative of the “inside” function.   This pattern of the integrand is precisely the derivative of a composition of two functions , where  is the anti-derivative of .  Thus, we get , where  and .  I used the example of  to illustrate.  This leads to the next step.  Another integration technique derived from a rule of differentiation is the technique called “Integration by Parts”.  Recall the Product Rulein differentiation:   (Note that this implies that the derivative of a product is NOT the product of the derivatives!)  Since each side of the equality is a function of , the respective anti-derivatives will also be equal (up to a constant!)  Thus we get a new relationship by integrating:  .  We can use the Fundamental Theorem of Calculus on the left hand side, and solve for one of the right hand side elements to get what is commonly called Integration by Parts rule, or what I call the Anti-product Rule:  .  Why this may create an advantage can almost be immediately seen.  Suppose you had a product of functions as an integrand.  Since it is obvious that the anti-derivative of a product will not in general be a product of the anti-derivatives, often it is not easy to see the anti-derivative readily.  But if the two factors are such that 1) the antiderivative of  is easy to calculate, while the derivative of  is easier to work with in an integral, then the right-hand side would be easier to calculate than the original problem.  I ended the lecture with the example , and a quick idea of just how this technique would work well here.

·       Wednesday, February 2:   I started this lecture with a review of the last lecture, noting that there is another notational way to see the Anti-Product Rule:  Namely, let  and , so  and .  Then we can write the rule as .  This makes the rule more of a double substitution.  However, I caution that it is not really a double substitution in the same way as that of the Substitution Rule, since one would not use this double sub to change the limits.  The limits stay the same in a definite integral under this new rule.  For examples, I did in detail the example I ended with last time, that of .  Then I did an example of a definite integral;  .  This last example provides another use for this technique:  When you find yourself with an integrand that is only a single function, and is one whose derivative you know and can work with in an integral.  In this case, let  be the entire integrand, and .  Then the right hand side, in many cases, becomes a readily computable integral.  In the case of the last example, we got  .  You can finish this one, no?  I then detailed the example of .  In this case, again let  and .  One can calculate  using a triangle and implicit differentiation (I did this in class).  Then using Integration by Parts, one gets , which can now be solved using the substitution  (do this in your notes!).  I also talked about the notion that sometimes you have to do this technique more than once, like in the example .  Here, let  and .  Then .  The right-hand-side integral requires another dose of Integration by parts.  The full calculation is .  I then talked about how sometimes, the purpose is not completely to “solve” an integral by integration, really.  It is more to “process” the integral to a point where one can “solve” for it algebraically.  I ended the lecture with the example of .  It is in the book, and I did the first part explicitly in class.  I chose my functions in class differently from the book, however, to give you an idea that one can perform these calculations differently from other people.  As long as it is done correctly, it does not matter.  For class, I chose  and  (compare with the example of the book).  I got .  While this seems to take us back to where we started from, we can now take the last equation and solve for , and wind up with , so that .  You should check that the derivative of this right hand side gives you the original integrand.

·       Friday, February 4:  Today, I finished the discussion on the Integration by Parts section with the example of , where of course  is an integer.  Letting, in this case,  and , one can use the IbP technique to show that the following formula holds:  , which I did in detail.  This is an example of a “reduction” formula, as an application of this technique reduces the degree of the resulting integral by 2.  One can see immediately that it would allow for the calculation of any positive power , although large powers would result in long calculations.  See the Extra Problem from Section 7.2 on the course webpage for an example.  Now, as an alternative, if  is odd, one could strip off a single factor , rewrite the other even powered , and with the substitution , we get .  While this does not look any easier, the integrand is really only a polynomial.  Multiply it out and the integration is easy.  Now, for many integrals involving trigonometric functions, a clever substitution or the use of a identity can be vital.  Section 7.2 is simply a section devoted to some patterns found in many integrals involving trigonometric functions, and some techniques for exposing their structure and solving them.  One pattern is .  If one of the powers is odd, then a quick substitution is very useful.  For example, if  is odd, then .  The last is simply a polynomial which can be multiplied out and integrated nicely.  A similar situation would hold if  were odd.  If they are both even, then the situation is less straightforward.  However, the Example for Section 7.2 that I drew up here is a nice example of what happens in this case.  I also talked about other patterns with some examples, like , or , where  and  are different (use an identity like  ).  This section is really just a collection of things like this.  Next class, we will move on to a more coherent technique involving integrals that are not trigonometric until AFTER the substitution.