Wi-fi Problem

 

Before Solve this problem, we have to understand some characteristics of odd functions.

Definition :Odd functions
A function f is said to be an odd function if `-f(x) = f(-x)`, for all value of `x`.

Observations: The graph of an odd function will be symmetrical about the origin.

Example: `f(x)=x^3` is odd
proof: `f(-x)=(-x)^3=-(x^3)=-f(x)`

Observe that the graph of `x^3` function is symmetrical about the origin. Now Let's talk about the integrals of odd functions.

Theorem
Let  `f`  be an integrable function on some closed interval that is symmetric about zero. Let's call the interval `[-a,a]` for `a>0`. If `f` is odd function then,

`\int_{-a}^{a}f(x)dx=0`

proof Let  `f`  be an integrable function on some closed interval that is symmetric about zero. Let's call the interval `[-a,a]` for `a>0`. Suppose that `f` is odd function.  Then, 

`\int_{-a}^{a}f(x)dx =\int_{-a}^{0}f(x)dx + \int_{0}^{a}f(x)dx-------(1)`

Before move on further,
Claim: `\int_{x=-a}^{0}f(x)dx = - \int_{x=0}^{a}f(x)dx`
proof of claim
Since `f` is odd, `f(-x)=-f(x)` Then,

`\int_{x=-a}^{0}f(x)dx = - \int_{x=a}^{0}f(-x)dx`

Let `u=−x, du=−dx`, then,

`-\int_{x=-a}^{0}f(-x)dx =\int_{u=a}^{0}f(u)du`

Reversing the limits of integration inverts the result, so

`\int_{u=a}^{0}f(u)du=-\int_{u=0}^{a}f(u)du`

Thus,

`\int_{x=-a}^{0}f(x)dx = -\int_{x=-a}^{0}f(-x)dx =\int_{u=a}^{0}f(u)du=-\int_{u=0}^{a}f(u)du=-\int_{x=0}^{a}f(x)dx`

This proove the claim,

By (1),

`\int_{-a}^{a}f(x)dx =\int_{-a}^{0}f(x)dx + \int_{0}^{a}f(x)dx` 
`\int_{-a}^{a}f(x)dx =-\int_{0}^{a}f(x)dx + \int_{0}^{a}f(x)dx=0`

Now let's move back to our problem. 

Integral `=\int_{-2}^{2}\left( x^3\cos(\frac{x}{2})+\frac{1}{2}\right)\sqrt(4-x^2)dx`

`=\int_{-2}^{2}\left( x^3\cos(\frac{x}{2})\right)\sqrt(4-x^2)dx+\int_{-2}^{2}\frac{1}{2}\sqrt(4-x^2)dx`

Let's call `h(x)= x^3\cos(\frac{x}{2})\sqrt(4-x^2)`. Then `h(x)` is odd. Because, `h(-x)=(-x)^3\cos(\frac{(-x)}{2})\sqrt(4-(-x)^2)=-x^3\cos(\frac{x}{2})\sqrt(4-x^2)=-h(x)`.

Then,

Integral `=\underbrace{\int_{-2}^{2}\left( x^3\cos(\frac{x}{2})\right)\sqrt(4-x^2)dx}_0 + \int_{-2}^{2}\frac{1}{2}\sqrt(4-x^2)dx`

Integral `=\int_{-2}^{2}\frac{1}{2}\sqrt(4-x^2)dx`

Integral `=\frac{1}{2}\int_{-2}^{2}\sqrt(4-x^2)dx`

Note that `x^2+y^2=4` is circle centerd at `(0,0)` with radius 2.  Hence, `\frac{1}{2}\int_{-2}^{2}\sqrt(4-x^2)dx` is area of semi circle with radius 2. Then, 

Integral `=\frac{1}{2}\frac{1}{2} \pi (2)^2dx=\pi\approx =3.1415926535`

Therefore answer,

`\therefore` Wi-Fi Password =31415926535