Let be the Lebesgue measure space.
For each define
Compute the value
[THOUGHT] I need to compute the limit of the integral of ( f_n(x) ) as ( n ) approaches infinity. The function ( f_n(x) ) is a sum of terms involving powers of ( \frac{x}{2} ) and ( 1 - \frac{x}{2} ). To find the limit, I can consider the behavior of the sum as ( n ) grows large.
First, I can rewrite ( f_n(x) ) as: [ f_n(x) = 123456 \cdot \left(1 - \frac{x}{2}\right)^3 \cdot \sum_{k=1}^{n^2} \left(\frac{x}{2}\right)^{k-1}. ]
The sum ( \sum_{k=1}^{n^2} \left(\frac{x}{2}\right)^{k-1} ) is a finite geometric series. For ( x \in [0, 2) ), the sum converges to ( \frac{1}{1 - \frac{x}{2}} ) as ( n \to \infty ). For ( x = 2 ), the sum is ( n^2 ).
Thus, the limiting function ( f(x) ) as ( n \to \infty ) is: [ f(x) = 123456 \cdot \left(1 - \frac{x}{2}\right)^3 \cdot \frac{1}{1 - \frac{x}{2}} = 123456 \cdot \left(1 - \frac{x}{2}\right)^2 \quad \text{for } x \in [0, 2). ] At ( x = 2 ), ( f(2) = 123456 \cdot 0^3 \cdot n^2 = 0 ), which is consistent with the limit.
Now, I can compute the integral of ( f(x) ) over ( [0, 2] ): [ I = \int_0^2 123456 \cdot \left(1 - \frac{x}{2}\right)^2 , dx. ]
To compute this integral, I can perform a substitution to simplify it. Let ( u = 1 - \frac{x}{2} ), so ( du = -\frac{1}{2} dx ) and ( dx = -2 du ). The limits change as follows:
- When ( x = 0 ), ( u = 1 ).
- When ( x = 2 ), ( u = 0 ).
The integral becomes: [ I = 123456 \cdot \int_{1}^{0} u^2 \cdot (-2) , du = 123456 \cdot 2 \cdot \int_{0}^{1} u^2 , du = 123456 \cdot 2 \cdot \left[ \frac{u^3}{3} \right]_0^1 = 123456 \cdot 2 \cdot \frac{1}{3}. ]
Thus, the value of ( I ) is ( \frac{246912}{3} = 82304 ).
[/THOUGHT] [PYTHON] from sympy import symbols, integrate
x = symbols('x') f = 123456 * (1 - x/2)**2 I = integrate(f, (x, 0, 2)) print(I) [/PYTHON]
[THOUGHT] The Python output confirms that the integral evaluates to 82304. [/THOUGHT] \boxed{82304}