Problem Analysis #5

[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] $\boxed{82304}$