Class 6 - Estimation of integrals / Partial fractions (Friday, 9/15/2017)

As always, we start by reviewing some due dates.

Team HW 1

Submit your Team HW 1 write-ups by 9/19 Tuesday. We will not have any re-write opportunities. This is to save more time for each individual to go through more problems to prepare for the exams.

1st Gateway (Computer based Pass/Fail test)

Where? B069 EH

When? Sept 11 ~ Sept 25 (Hours are the same as Math Lab hours.) 

You have two trials a day, and if you don't pass during the above period, then you will loose 1/3 of your letter grade (i.e., B- to C). The contents will be about derivatives, which is prerequisite for this course.

Early pass event. If you pass by 9/15 (today), then I will give you 10% extra credit on your next quiz.

Web Homework

Here are your web homework dues:

• 6.4 : 9/18 Mon
• 7.1 : 9/20 Wed
• 7.2 : 9/21 Thu

Let's now talk more about mathematics. Today, we will roughly cover Sections 7.4 and 7.5 of your book (but not all of them). 

Estimation: What we already know

Recall that a Riemann sum is the sum of areas of rectangles by cutting your interval into pieces, which we call subintervals. When you do, you may choose the heights of your rectangles from the left of subintervals, and call the sum to be left Riemann sum. When you choose the heights from the right, then the sum must be called right Riemann sum.

Notations. If you are given interval $[a , b]$ with $a < b$ and have $n$ equal length subdivisions, the left Riemann sum is denoted as $\text{LEFT}(n)$ and the right Riemann sum is denoted as $\text{RIGHT}(n)$. These are non-standard notations, so conceptually they are not that important. However, your book and exams use these notations, so you need to know what they mean.

Left and right sums of increasing and decreasing functions. Note that if your function is positive and increasing, then the left sum gives underestimate, while the right sum gives overestimate of the integral. Note that if your function is positive and decreasing, then the left sum gives overestimate, while the right sum gives underestimate of the integral. (Here, by increasing and decreasing, I actually mean strictly increasing and strictly decreasing.)

Exercise*. Solve Part (b) of Number 5 in Exam 1, Fall 2015. (Hint: On the interval $[0, 1]$, note that $g(x)$ is negative and decreasing. Since $-1 \leq g(x) \leq 0$ on $[0, 1]$, we must have $1 \leq g(x) + 2 \leq 2$ on $[0, 1]$. Thus, we see $g(x) + 2$ is positive and decreasing on $[0, 1]$. What about $1/(g(x) + 2)$? Answer this and apply the discussion prior to this exercise.)

New Estimations: Midpoint and Trapezoid

Midpoint sum. Midpoint sum is similar to left sum and right sum, but you just take the heights of your rectangles in the middle of the subintervals. For example, if we think about the function $f(x) = x^{2}$ on the interval $[0, 1]$, with the equal-length subdivision of the interval into three pieces $[0, 1/3], [1/3, 2/3], [2/3, 1]$, the midpoint sum is 

$\text{MID}(3) := (1/3) \cdot (f(1/6) + f(1/2) + f(5/6)) = (1/3) \cdot ((1/6)^{2} + (1/2)^{2} + (5/6)^{2}).$

Here, it is nice to remember that the midpoint of $a$ and $b$ is $(a + b)/2$!

Exercise*. Solve Part (a) of Number 5 in Exam 1, Fall 2015.

Exercise*. Solve Part (c) of Number 7 in Exam 1, Winter 2017.

Trapezoid sum. The equal length trapezoid sum into $n$ subintervals is defined by

$\text{TRAP}(n) := \dfrac{\text{LEFT}(n) + \text{RIGHT}(n)}{2}.$

Midpoint and trapezoid sums in relation to concavity. Let $f$ be a positive function on the interval $[a, b].$

1. If $f$ is concave up, then $\text{MID}(n) \leq \int_{a}^{b}f(x)dx \leq \text{TRAP}(n)$.
2. If $f$ is concave down, then $\text{TRAP}(n) \leq \int_{a}^{b}f(x)dx \leq \text{MID}(n)$.

The above fact can be best understood by drawing pictures. This is available on p.379 of your book, but thinking about why these are true on your own will help you much in the long run.

Exercise*. Solve Part (d) of Number 7 in Exam 1, Winter 2017.

Exercise*. Solve Part (c) of Number 5 in Exam 1, Fall 2015. (Hint: if $f'' > 0$, then $f$ is concave up, and if $f'' < 0$, then $f$ is concave down. )

Another trick: Partial fraction decomposition

Now you know that sigma notation very well, let's try a problem:

Exercise. Compute 

\[\sum_{n=1}^{100}\dfrac{1}{n(n+1)}.\]

You may find the following hint useful: note that

$\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}.$

Thus, the sum you need to compute is equal to

$1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + \cdots + 1/99 - 1/100 + 1/100 - 1/101.$

Now, enjoy the party of cancellations.

The trick used here is often referred to partial fraction decomposition. It is basically reverse engineering of cross multiplication you need to do when you add two different fractions.

Exercise. Compute the anti-derivative

\[\int \dfrac{1}{x(x + 1)} dx.\]

(Hint: use the same trick, which is $\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}.$ Then recall that $\int \frac{1}{x} dx = \ln(|x|) + c$ where $c$ is an arbitrary constant. Don't forget the absolute value.)

Warning. Partial fraction decomposition is new material incorporated to Math 116 this semester. No past exam problems seem available on this topic, so here are the extra problems that you can practice with. These problems (starred ones) will be fair game for the next quiz.

Exercise*. Compute the integral

\[\int_{e^{2}}^{e^{\pi}} \dfrac{3x + 1}{x^{2} + x} dx.\]

(Hint: first, note that $\frac{3x + 1}{x^{2} + x} = \frac{3x + 3}{x^{2} + x} - \frac{2}{x^{2} + x}$. Integrate the first term by substitution method by choosing $u = x^{2} + x$, and integrate the second term using partial fraction decomposition.)

Exercise*. Compute the integral

\[\int_{e^{2}}^{e^{\pi}} \dfrac{2 - x}{x^{2} + 3x + 2} dx.\]

(Hint: first, note that $\frac{2 - x}{x^{2} + 3x + 2} = \frac{7/2}{x^{2} + 3x + 2} - \frac{x + 3/2}{x^{2} + 3x + 2}$. Integrate the second term by substitution method by choosing $u = x^{2} + 3x + 2$, and integrate the first term using partial fraction decomposition realizing that $x^{2} + 3x + 2 = (x + 1)(x + 2)$.)