Class 3 - Two tricks to compute integrals (Friday, 9/8/2017)

Before we talk about mathematics, here is important logistics:

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, then I will give you 10% extra credit on your next quiz.

Web Homework

Here are your web homework dues:

• 5.1, 5.2, 5.3 : 9/11 Mon
• 5.4, 6.1, 6.2 : 9/13 Wed

1st Quiz on Tuesday

We will have the first quiz on Tuesday (9/12/2017):

• How long? I will give you 50 minutes to finish the quiz, and you may leave early if you are done solving the problems and checking your solutions.
• On what? All the started exercises (i.e., "Exercise*") in the notes of Class 1, 2, and 3 (the last of which is this one).
• Before the quiz? For about 30 minutes, I can answer your questions and go over some problems we did not because we were short on time.
• What to write? You cannot just write answers. If you just write down the answers without neatly writing out the process of getting them, you will only obtain 1 point per correct answer. This is because on uniform exams, much of the grading of each problem will be based on how you get to the correct answer, on top of the answer itself.

Let's get back to where we were at now:

Review: Fundamental Theorems

So far, we have learned two fundamental theorems:

1st Fundamental Theorem of Calculus. For any function $f$ with continuous derivative $f'$, we have
\[\int_{a}^{b}f'(x)dx = f(b) - f(a),\]
where $a, b$ are any fixed real numbers.

2nd Fundamental Theorem of Calculus. For any continuous function $f$, we have
\[\frac{d}{dx}\int_{a}^{x}f(t)dt = f(x),\]
where $a$ is any fixed real number.

Exercise. What are the anti-derivatives of the following functions?

1. $f(x) = \sqrt{2} x \mapsto x^{3} + \pi x + e$;
2. $f(x) = \sin(x/2)$;
3. $f(x) = \cos(2x)$;
4. $f(x) = \sec^{2}(2x) = 1/\cos^{2}(2x)$;
5. $f(x) = e^{x}$;
6. $f(x) = 2^{x}$;
7. $f(x) = 1/(e\pi x)$.

(Hints: For Part 4, recall that $\frac{d}{dt}\tan(t) = \sec^{2}(t)$. For Part 3, recall that $\frac{d}{dt}\ln(t) = 1/t$ when $t > 0$, but also note that when $t < 0$, we have $\frac{d}{dt}\ln(-t) = 1/t$. Don't forget to write out the arbitrary constant.)

Exercise*. Solve Parts (a) and (b) of Number 7 of Exam 1, Winter 2017. (Hints: for Part (b)-(i), it helps to write $F(x) = H(x^{2})$ where $H(y) = \int_{7}^{y}e^{-t^{2}}dt$. By chain rule, note that $F'(x) = 2xH'(x^{2})$. Now, you just need to compute $H'(y)$ and then take $y = x^{2}$.)

We will roughly cover Section 7.1 and 7.2 of your book today. Again, recall that we started off by defining integral by area under the graph. The fundamental theorems above tell us that you can do the reverse-engineering of taking derivative (i.e., take anti-derivative) to compute the integral (i.e., the area). Hence, if we know something about derivative, we should be able to use it for computing integrals. We had chain rule and product rule when we take derivatives, so these give two tricks to compute integrals.

Trick 1: Reverse chain rule a.k.a "Integration by Substitution"

Chain rule states: $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x).$

Taking integral operators on both sides, we get

\[\int_{a}^{b}\frac{d}{dx}f(g(x))dx = \int_{a}^{b}f'(g(x))g'(x)dx.\]

By the 1st fundamental theorem, the left-hand side is equal to $f(g(b)) - f(g(a))$, we we get:

Substitution trick. We have

\[\int_{a}^{b}f'(g(x))g'(x)dx = f(g(b)) - f(g(a)).\]

Exercise*. Solve Part (b) of Number 1 in Exam 1, Fall 2016. (Hint: Take $g(x) = x^{2}$ in substitution method described above. You will have to divide by $2$ after that.)

Another way to write integration by substitution. Using 1st fundamental theorem, we may rewrite above as

\[\int_{a}^{b}f'(g(x))g'(x)dx = \int_{g(a)}^{g(b)}f'(u)du.\]

Exercise*. Solve Number 7 in Exam 1, Fall 2016. (Hint: By 2nd fundamental theorem, we have $G'(x) = g(x)$. Hence, if we write $u = G(x)$, then $du = g'(x)du$. The integral in question then simplifies to $\int_{G(0)}^{G(7)}u^{2}du$.)

Exercise*. Solve Number 2 in Exam 1, Fall 2016. (Hint: You have $S(1) = 2$, so $S(t) = 2 + \int_{1}^{t}S'(x)dx.$)

Trick 2: Reverse product rule a.k.a "Integration by Parts"

Product rule states: $\frac{d}{dx}u(x)v(x) = u'(x)v(x) + u(x)v'(x).$

Taking integral operators on both sides, we get

\[\int_{a}^{b}\frac{d}{dx}u(x)v(x)dx = \int_{a}^{b}u'(x)v(x)dx + \int u(x)v'(x)dx.\]

By the 1st fundamental theorem, the left-hand side is equal to $f(b)g(b) - f(a)g(a)$. Hence, we get:

Integration by parts. We have

\[\int_{a}^{b}u'(x)v(x)dx + \int_{a}^{b}u(x)v'(x)dx = u(b)v(b) - u(a)v(a).\]

Exercise. Compute the integral

\[\int_{1}^{e^{10}} \ln(x) dx.\]

(Hint: In the statement s of the integration by parts above, take $x = f(x)$ and $g(x) = \ln(x)$. Then, you get $\int_{1}^{e^{10}}\ln(x)dx + \int_{1}^{e^{10}}1dx = e^{10}\ln(e^{10}).$)

Exercise*. Solve Part (a) of Number 1 in Exam 1, Fall 2016. (Hint: Differentiate $xf(x)$ in $x$ and see what you get!)

Another way to write integration by parts. Moving a term to the other side, we may rewrite the formula above as
\[\int_{a}^{b}u(x)v'(x)dx = u(x)v(x)|_{a}^{b} - \int_{a}^{b}u'(x)v(x)dx.\]

Exercise*. Solve Part (c) of Number 1 in Exam 1, Fall 2016. (Hint: Take $u(x) = x^{2}/2$ and $v'(x) = 2xf'(x^{2})$ in the above formula. Apply the formula and then use the substitution method.)