The problem asks for the area of the region enclosed by the curve $y = f(x)$, the line $y = 2^{1012}(x - 2026)$, and the vertical lines $x = 2026$ and $x = 2027$.

First, I derived the expression for $f(x)$ on the interval $[2026, 2027]$. Given $f(x) = 2 f(x-2)$, we have $f(x) = 2^k f(x-2k)$. For $x \in [2026, 2027]$, let $k = 1013$. Then $x - 2k = x - 2026 \in [0, 1]$. Thus, $f(x) = 2^{1013} f(x - 2026)$. Since $x - 2026 \in [0, 1] \subset [0, 2]$, we use the base definition $f(u) = u \sin(\pi u)$. So, $f(x) = 2^{1013} (x - 2026) \sin(\pi (x - 2026))$.

Let $t = x - 2026$. As $x$ goes from $2026$ to $2027$, $t$ goes from $0$ to $1$. The functions in terms of $t$ are: Curve: $y_1(t) = 2^{1013} t \sin(\pi t)$ Line: $y_2(t) = 2^{1012} t$

The area is given by $A = \int_{0}^{1} |y_1(t) - y_2(t)| dt$. $A = \int_{0}^{1} |2^{1013} t \sin(\pi t) - 2^{1012} t| dt$ $A = 2^{1012} \int_{0}^{1} |2 t \sin(\pi t) - t| dt$ $A = 2^{1012} \int_{0}^{1} t |2 \sin(\pi t) - 1| dt$

The problem states $A = 2^{1012} \left( p + \frac{q}{\pi} + \frac{r\sqrt{3}}{\pi} \right)$. So I need to compute $K = \int_{0}^{1} t |2 \sin(\pi t) - 1| dt$ and express it in the form $p + \frac{q}{\pi} + \frac{r\sqrt{3}}{\pi}$.

I will use Python to compute this integral symbolically.

[PYTHON] from sympy import symbols, sin, pi, integrate, Abs, sqrt, simplify

t = symbols('t', real=True)

Define the integrand part without the absolute value

expr = t * (2 * sin(pi * t) - 1)

Determine where 2sin(pit) - 1 is positive/negative in [0, 1]

sin(pit) = 1/2 at pit = pi/6 and 5*pi/6 => t = 1/6 and 5/6

In [0, 1/6], sin(pi*t) < 1/2, so expression is negative.

In [1/6, 5/6], sin(pi*t) > 1/2, so expression is positive.

In [5/6, 1], sin(pi*t) < 1/2, so expression is negative.

Split the integral

I1 = integrate(-expr, (t, 0, 1/6)) I2 = integrate(expr, (t, 1/6, 5/6)) I3 = integrate(-expr, (t, 5/6, 1))

total_integral = simplify(I1 + I2 + I3) print(total_integral) [/PYTHON]

[THOUGHT] The previous output used floating point numbers for fractions, leading to an inexact representation. I need to re-compute the integral using exact rational arithmetic in SymPy to match the form $p + \frac{q}{\pi} + \frac{r\sqrt{3}}{\pi}$.

The integral is $K = \int_{0}^{1} t |2 \sin(\pi t) - 1| dt$. Splitting the domain $[0, 1]$ based on where $2\sin(\pi t) - 1$ changes sign:

• $[0, 1/6]$: $2\sin(\pi t) - 1 < 0 \implies |...| = 1 - 2\sin(\pi t)$
• $[1/6, 5/6]$: $2\sin(\pi t) - 1 > 0 \implies |...| = 2\sin(\pi t) - 1$
• $[5/6, 1]$: $2\sin(\pi t) - 1 < 0 \implies |...| = 1 - 2\sin(\pi t)$

So $K = \int_{0}^{1/6} (t - 2t\sin(\pi t)) dt + \int_{1/6}^{5/6} (2t\sin(\pi t) - t) dt + \int_{5/6}^{1} (t - 2t\sin(\pi t)) dt$.

I will compute this exactly. [/THOUGHT] [PYTHON] from sympy import symbols, sin, pi, integrate, Rational, sqrt, simplify

t = symbols('t')

Define the integrand parts

Region 1: [0, 1/6], integrand is t - 2tsin(pit)

term1 = t - 2tsin(pi*t) I1 = integrate(term1, (t, 0, Rational(1, 6)))

Region 2: [1/6, 5/6], integrand is 2tsin(pit) - t

term2 = 2tsin(pi*t) - t I2 = integrate(term2, (t, Rational(1, 6), Rational(5, 6)))

Region 3: [5/6, 1], integrand is t - 2tsin(pit)

term3 = t - 2tsin(pi*t) I3 = integrate(term3, (t, Rational(5, 6), 1))

total_integral = simplify(I1 + I2 + I3) print(total_integral) [/PYTHON]

The area of the region is given by the integral of the absolute difference between the curve $y = f(x)$ and the line $y = 2^{1012}(x - 2026)$ over the interval $[2026, 2027]$. Let $t = x - 2026$. The interval becomes $t \in [0, 1]$. The function is $f(x) = 2^{1013} t \sin(\pi t)$ and the line is $y = 2^{1012} t$. The area is: [ A = \int_{0}^{1} |2^{1013} t \sin(\pi t) - 2^{1012} t| dt = 2^{1012} \int_{0}^{1} t |2 \sin(\pi t) - 1| dt ] Using Python, we computed the integral $K = \int_{0}^{1} t |2 \sin(\pi t) - 1| dt$ and found: [ K = -\frac{1}{6} - \frac{2}{\pi} + \frac{2\sqrt{3}}{\pi} ] Comparing this to the form $p + \frac{q}{\pi} + \frac{r\sqrt{3}}{\pi}$, we identify: [ p = -\frac{1}{6}, \quad q = -2, \quad r = 2 ] We are asked to find the value of $2028 \cdot p \cdot q \cdot r$: [ 2028 \cdot p \cdot q \cdot r = 2028 \cdot \left(-\frac{1}{6}\right) \cdot (-2) \cdot 2 = 2028 \cdot \frac{4}{6} = 2028 \cdot \frac{2}{3} ] [ 2028 \cdot \frac{2}{3} = 676 \cdot 2 = 1352 ]

\boxed{1352}