China 南京航空航天大学 Nanjing University of Aeronautics and Astronautics set the WiFi password as the answer of this integral (first 6 digits).

Can you solve it?

(If can’t, please revise GCE “A-level” / Baccalaureate / 高考 Calculus 微积分)

Answer : Break the integral (I) into 2 parts:

I = A(x) + B(x)

$\displaystyle A(x) = \int_{-2 }^ {2} x^{3}. \cos \frac{x}{2}.\sqrt{4-x^2}dx$

$\displaystyle B(x) = \int_{-2 }^ {2} \frac{1}{2}\sqrt{4-x^2}dx$

A(x) = – A(-x) => Odd function
=> A(x) = 0 since its area canceled out over [-2, 2]

B(x) = B(-x) => Even function
$\displaystyle\implies B(x) = 2\int_{0 }^ {2} \frac{1}{2}\sqrt{4 - x^2}dx$
$\displaystyle\implies B(x) = \int_{0 }^ {2} \sqrt{4 - x^2}dx$

Let x = 2 sin t => dx = 2 cos t. dt

x = 2 = 2 sin t => sin t = 1 => t = π / 2

x = 0 = 2 sin t => sin t = 0 => t = 0

$\displaystyle B(x) = \int_{0 }^ {\pi/2} \sqrt{4 - 4.\sin^{2} {t} }. (2 \cos t. dt)$

$\displaystyle \implies B(x) = \int_{0 }^ {\pi/2} 2.\cos t. (2 \cos t. dt)$

$\displaystyle\implies B(x) = \int_{0 }^ {\pi/2} 4 \cos^{2} t. dt$

$\displaystyle \cos ^{2} t = \frac {1 + \cos 2t} {2 }$

$\displaystyle\implies B(x) = \int_{0 }^ {\pi/2} (2 + 2\cos 2t) . dt$

$\displaystyle\implies B(x) = (2 t) \Bigr|_{0 }^ {\pi/2} + (2. \frac{1}{2} \sin 2t) \Bigr|_{0 }^ {\pi/2}$

$\displaystyle\implies B(x) = (\pi ) + \sin \pi = \pi$

$\boxed{ I = \pi = 3.14159}$

A smarter method using Analytic Geometry: A circle of radius 2 is

$x^2 + y^2 = 4 \implies y = \sqrt {4 -x^2}$

# What is math?

What is Math ? Interesting article below:

https://infinityplusonemath.wordpress.com/2017/06/17/what-is-math/

• Mathematics = “that which is learned“ –(Pythagoras)

Math is not about calculation, it is understanding the nature, the universe, the philosophy (logic, intelligence – both “human” and “artificial”)…

What is Axiom, Lemma, Proposition ? Why rigorous Calculus was needed hundred years after Newton & Leibniz had invented it – “Epsilon-Delta” Analysis.

Difference between Riemann Integral & Lebesgue Integral ?

Calculus = 微积分

“尽小者大, 积微者著”
=> 見微知著

# Calculus Fundamental Technique

$D first, then \int$

This is just a simple but powerful application of Calculus, behind which lies the philosophy of Leibniz:

1. D (=dy/dx) is the inverse function of $\int$
2. Calculus Fundamental Technique: $D first, then \int$
E.g. Sherlock Holmes example:
1. D first:
dT/dt = k(T-Ts)
=> can’t solve directly
2. Take D’s inverse:
$\int {dT/(T-Ts)} = k.dt$
=> can solve now !

# Solution: Sherlock Holmes

Apply Newton’s Law of Cooling:
$dT/dt = k(T-T_s)$
$T_s =21$ (Room Temperature)
$dT/(T-T_s)= k.dt$
$\int \frac {dT}{T-T_s}=\int k.\mathrm{d}t$
$ln (T-T_s)=kt+C$
$e^{kt+C}= T-T_s$
$T= Ae^{kt} +T_s$
$A=e^{C}$
$T=Ae^{kt}+21$

At t=0, (normal body temperature)
$T_o= A + 21 =37$
=> A=16

Let t =x hour 1st temperature taken
$T(x)= 16e^{kx}+21=29$
$16e^{kx} = 8$
$e^{kx}= \frac {1}{2}$ …[1]

t = x+1 hour later

$T(x+1) =16e^{k(x+1)} + 21 = 27$
$e^{k(x+1)} = 3/8$
$e^{kx}e^{k}= 3/8$
$e^{k} = 3/4$
k = ln(3/4)

[1]: kx = ln(1/2)
x = ln(1/2) / ln(3/4)
x=2.41 hr = 2 hr 25m
Murder Time= 2am -2h25m =11:35 p.m. [QED]

# Newtonian Calculus not rigorous !

Why Newton’s Calculus Not Rigorous?

$f(x ) = \frac {x(x^2+ 5)} { x}$ …[1]

cancel x (≠0)from upper and below => $f(x )=x^2 +5$

$\mathop {\lim }\limits_{x \to 0} f(x) =x^2 +5= L=5$ …[2]

In [1]: we assume x ≠ 0, so cancel upper & lower x
But In [2]: assume x=0 to get L=5
[1] (x ≠ 0) contradicts with [2] (x = 0)

This is the weakness of Newtonian Calculus, made rigorous later by Cauchy’s ε-δ ‘Analysis’.