[THOUGHT] I'll compute the area as an integral using the periodic definition of (f(x)), reduce it to an integral over (u \in [0,1]), evaluate the integral symbolically with Sympy, then extract the rational coefficients (p,q,r). Finally I'll compute (2028 \cdot p \cdot q \cdot r) in Python. [/THOUGHT] [PYTHON] import sympy as sp

u = sp.symbols('u', real=True) expr = u*(2sp.sin(sp.piu) - 1)

Define the absolute-value integral I = ∫_0^1 |expr| du

I = -sp.integrate(expr, (u, 0, sp.Rational(1,6)))
+ sp.integrate(expr, (u, sp.Rational(1,6), sp.Rational(5,6)))
- sp.integrate(expr, (u, sp.Rational(5,6), 1))

I_simplified = sp.simplify(I) print("I =", I_simplified)

Express I in form p + q/pi + r*sqrt(3)/pi

p, q, r = sp.symbols('p q r', rational=True) pi = sp.pi sqrt3 = sp.sqrt(3)

Solve for p,q,r by matching coefficients

I = p + q/pi + r*sqrt3/pi

target = p + q/pi + r*sqrt3/pi eq = sp.Eq(sp.simplify(I_simplified - target), 0)

Multiply both sides by pi to clear denominators

eq_mul = sp.Eq(sp.simplify((I_simplified - target)*pi), 0) expr_coeff = sp.simplify((I_simplified - target)*pi)

Separate into a + b*sqrt(3) form and match

A = sp.simplify(sp.nsimplify(sp.expand(expr_coeff), [sqrt3]))

Try to express as a + b*sqrt(3) with a,b in Q

a = sp.simplify(sp.expand(A).coeff(sqrt3, 0)) b = sp.simplify(sp.expand(A).coeff(sqrt3, 1))

print("expr_coeff =", expr_coeff) print("a (no sqrt3) =", a) print("b (sqrt3 coeff) =", b)

We want expr_coeff = 0 => a + b*sqrt(3) = 0

But in terms of p,q,r: expr_coeff = (I - p - q/pi - r*sqrt3/pi)*pi

Let's instead directly solve for p,q,r by equating coefficients:

p_var, q_var, r_var = sp.symbols('p_var q_var r_var')

target2 = p_var + q_var/pi + r_var*sqrt3/pi expr2 = sp.simplify(I_simplified - target2)

Multiply by pi and separate constants and sqrt3

expr2_pi = sp.simplify(expr2 * pi) const_part = sp.simplify(sp.expand(expr2_pi).coeff(sqrt3, 0)) sqrt3_part = sp.simplify(sp.expand(expr2_pi).coeff(sqrt3, 1))

const_part = 0 and sqrt3_part = 0 as polynomials in 1 and 1/pi.

Write const_part as A + B/pi

A_const = sp.simplify(sp.nsimplify(const_part, [pi])) B_const = sp.simplify(sp.expand(const_part - A_const)*pi) # rough attempt

print("I_simplified exact:", I_simplified) print("Numeric check of I:", float(I_simplified))

Alternatively, directly integrate symbolically without abs then split manually

Let's instead compute symbolic pieces separately and simplify manually:

int1 = sp.integrate(expr, (u, 0, sp.Rational(1,6))) int2 = sp.integrate(expr, (u, sp.Rational(1,6), sp.Rational(5,6))) int3 = sp.integrate(expr, (u, sp.Rational(5,6), 1))

A1 = -int1 A2 = int2 A3 = -int3

print("int1:", sp.simplify(int1)) print("int2:", sp.simplify(int2)) print("int3:", sp.simplify(int3)) print("A1:", sp.simplify(A1)) print("A2:", sp.simplify(A2)) print("A3:", sp.simplify(A3))

print("Check sum A1+A2+A3:", sp.simplify(A1+A2+A3))

Now express A1+A2+A3 explicitly in desired form by nsimplify with basis [1, 1/pi, sqrt(3)/pi]

basis = [1, 1/sp.pi, sp.sqrt(3)/sp.pi] coeffs = sp.nsimplify(I_simplified, basis) print("nsimplify coefficients as combination of [1,1/pi,sqrt(3)/pi]:", coeffs)

Extract p,q,r from that combination

coeffs should be a11 + a2(1/pi) + a3*(sqrt(3)/pi)

a1, a2, a3 = coeffs.as_coefficients_dict()[1], coeffs.as_coefficients_dict()[1/sp.pi], coeffs.as_coefficients_dict()[sp.sqrt(3)/sp.pi] print("p,q,r:", a1, a2, a3)

Now compute 2028 * p * q * r

value = 2028 * a1 * a2 * a3 print("2028 * p * q * r =", sp.simplify(value)) [/PYTHON]

[THOUGHT] From the Python output, the integral value is [ I = \frac{-12 - \pi + 12\sqrt{3}}{6\pi}. ] Rewrite: [ I = \frac{-12 - \pi}{6\pi} + \frac{12\sqrt{3}}{6\pi} = -\frac{12}{6\pi} - \frac{\pi}{6\pi} + \frac{12}{6}\frac{\sqrt{3}}{\pi} = -\frac{2}{\pi} - \frac{1}{6} + 2\frac{\sqrt{3}}{\pi}. ] Thus [ I = p + \frac{q}{\pi} + \frac{r\sqrt{3}}{\pi} ] with [ p = -\frac{1}{6},\quad q = -2,\quad r = 2. ] Hence [ 2028 \cdot p \cdot q \cdot r = 2028 \cdot \left(-\frac{1}{6}\right)\cdot(-2)\cdot 2 = 1352. ] [/THOUGHT] \boxed{1352}