Problem Analysis #5

[THOUGHT] First, analyze the function $f_n(x) = \sum_{k=1}^{n^2} 123456 \left(\frac{x}{2}\right)^{k-1} \left(1 - \frac{x}{2}\right)^3$ for $x \in [0,2]$.

Factor out the constant and the term independent of $k$:
$f_n(x) = 123456 \left(1 - \frac{x}{2}\right)^3 \sum_{k=1}^{n^2} \left(\frac{x}{2}\right)^{k-1}$

The sum is a geometric series with ratio $r = \frac{x}{2}$, from $k=1$ to $k=n^2$. Let $j = k-1$, so the sum becomes:
$\sum_{j=0}^{n^2-1} \left(\frac{x}{2}\right)^j = \frac{1 - \left(\frac{x}{2}\right)^{n^2}}{1 - \frac{x}{2}} \quad \text{for} \quad x \neq 2$

Thus,
$f_n(x) = 123456 \left(1 - \frac{x}{2}\right)^3 \cdot \frac{1 - \left(\frac{x}{2}\right)^{n^2}}{1 - \frac{x}{2}} = 123456 \left(1 - \frac{x}{2}\right)^2 \left(1 - \left(\frac{x}{2}\right)^{n^2}\right)$

Now, compute the integral:
$I = \lim_{n \to \infty} \int_0^2 f_n(x) dx = \lim_{n \to \infty} 123456 \int_0^2 \left(1 - \frac{x}{2}\right)^2 \left(1 - \left(\frac{x}{2}\right)^{n^2}\right) dx$

Split the integral:
$I = 123456 \left[ \int_0^2 \left(1 - \frac{x}{2}\right)^2 dx - \int_0^2 \left(1 - \frac{x}{2}\right)^2 \left(\frac{x}{2}\right)^{n^2} dx \right]$

As $n \to \infty$, the second integral: for $x \in [0,2)$, $\left(\frac{x}{2}\right)^{n^2} \to 0$ because $\frac{x}{2} < 1$. At $x = 2$, $\left(\frac{x}{2}\right)^{n^2} = 1$, but the set $\{2\}$ has Lebesgue measure zero. By dominated convergence, the second integral vanishes:
$\lim_{n \to \infty} \int_0^2 \left(1 - \frac{x}{2}\right)^2 \left(\frac{x}{2}\right)^{n^2} dx = 0$

Thus,
$I = 123456 \int_0^2 \left(1 - \frac{x}{2}\right)^2 dx$

Compute the integral using substitution. Let $u = 1 - \frac{x}{2}$, so $du = -\frac{1}{2} dx$, and $dx = -2 du$. When $x = 0$, $u = 1$; when $x = 2$, $u = 0$:
$\int_0^2 \left(1 - \frac{x}{2}\right)^2 dx = \int_1^0 u^2 (-2) du = 2 \int_0^1 u^2 du = 2 \left[ \frac{u^3}{3} \right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3}$

Therefore,
$I = 123456 \cdot \frac{2}{3} = \frac{246912}{3} = 82304$

Verification: 123456 * 2 = 246912; 246912 / 3 = 82304.

Final answer:
$\boxed{82304}$

[THOUGHT] I need to compute the integral of (1 - x/2)^2 from 0 to 2 and then multiply by 123456 to verify the result. [/THOUGHT] [PYTHON] from scipy.integrate import quad

def g(x): return (1 - x/2)**2

result, error = quad(g, 0, 2) I = 123456 * result print(I) [/PYTHON]

$\boxed{82304}$