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 $x^2 + y^2 = 4 \implies y = \sqrt {4 -x^2}$ 1. tomcircle