China Nanjing Aviation University set the WiFi password as the answer of this integral (first 6 digits).

Can you solve it?

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

I = A(x) + B(x)

$latex A(x) = int_{-2 }^ {2} x^{3}. cos (x/2).sqrt{4-x^2}dx$

$latex B(x) = int_{-2 }^ {2} 1/2.sqrt{4-x^2}dx$

**A(x) = – A(-x) => Odd function **

=> **A =…** since its area canceled out over [-2, 2]

**B(x) = B(-x) => Even function**

$latex implies B(x) = 2.int_{0 }^ {2} 1/2.sqrt{4 – x^2}dx$

$latex implies B(x) = int_{0 }^ {2} sqrt{4 – x^2}dx$

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 =…

$latex B(x) = int_{0 }^ {pi/2} sqrt{4 – 4.sin^{2} {t}…

View original post 96 more words