Let a function defined on the set of all real numbers satisfyandConsider the region enclosed by the curve , the line and the two vertical lines and .The area of this region is equal tofor some rational numbers .Find the value of 2028 $$\cdot$$ p $$\cdot$$ q $$\cdot$$ r.oindent
[THOUGHT] The function f(x) for x in [2026, 2027] is f(x) = 2^{1013} t sin(π t) where t = x - 2026. The line is l(x) = 2^{1012} t. The area is 2^{1012} ∫_0^1 t |2 sin(π t) - 1| dt. The points where 2 sin(π t) = 1 are t=1/6 and t=5/6. Split the integral accordingly and compute exactly using sympy to verify p, q, r. [/THOUGHT] [PYTHON] import sympy as sp
t, pi = sp.symbols('t pi') a = sp.Rational(1,6) b = sp.Rational(5,6)
F = lambda u: - (u/pi) * sp.cos(piu) + (1/pi**2) * sp.sin(piu) G = lambda u: u**2 / 2
int1 = G(a) - G(0) - 2*(F(a) - F(0)) int2 = 2*(F(b) - F(a)) - (G(b) - G(a)) int3 = G(1) - G(b) - 2*(F(1) - F(b))
I = int1 + int2 + int3 print(I.simplify()) print(sp.N(I)) [/PYTHON]
[THOUGHT] The integral (I = -\frac{1}{6} - \frac{2}{\pi} + \frac{2\sqrt{3}}{\pi}), so (p = -\frac{1}{6}), (q = -2), (r = 2). Then (2028 \cdot p \cdot q \cdot r = 2028 \cdot \left(-\frac{1}{6}\right) \cdot (-2) \cdot 2 = 2028 \cdot \frac{2}{3} = 1352). [/THOUGHT] \boxed{1352}
Explanation
The reasoning derives from analyzing the function f(x) and the given boundaries. For x in [2026, 2027], let t = x - 2026, so t ∈ [0, 1]. Since 2026 = 2 × 1013, f(x) = 2^{1013} f(t) = 2^{1013} t sin(π t). The line is y = 2^{1012} t. The area is ∫_0^1 |2^{1013} t sin(π t) - 2^{1012} t| dt = 2^{1012} ∫_0^1 t |2 sin(π t) - 1| dt, denoted as 2^{1012} I.
The expression 2 sin(π t) - 1 changes sign where sin(π t) = 1/2, at t = 1/6 and t = 5/6. Thus, split I into three intervals: [0, 1/6], [1/6, 5/6], and [5/6, 1].
- For [0, 1/6] and [5/6, 1], 2 sin(π t) - 1 ≤ 0, so |2 sin(π t) - 1| = 1 - 2 sin(π t).
- For [1/6, 5/6], 2 sin(π t) - 1 ≥ 0, so |2 sin(π t) - 1| = 2 sin(π t) - 1.
Thus, I = ∫0^{1/6} t (1 - 2 sin(π t)) dt + ∫{1/6}^{5/6} t (2 sin(π t) - 1) dt + ∫_{5/6}^1 t (1 - 2 sin(π t)) dt.
The antiderivative of t sin(π t) is F(t) = - (t/π) cos(π t) + (1/π^2) sin(π t), and of t is G(t) = t^2 / 2.
Evaluate:
- int1 = [G(1/6) - G(0)] - 2 [F(1/6) - F(0)] = 1/72 + √3/(6 π) - 1/π^2.
- int2 = 2 [F(5/6) - F(1/6)] - [G(5/6) - G(1/6)] = √3 / π - 1/3.
- int3 = [G(1) - G(5/6)] - 2 [F(1) - F(5/6)] = 11/72 - 2/π + 5 √3 /(6 π) + 1/π^2.
Summing:
- Rational terms: 1/72 - 1/3 + 11/72 = -1/6.
- Coefficients of 1/π: -2.
- Coefficients of √3 / π: √3/(6 π) + √3 / π + 5 √3 /(6 π) = 2 √3 / π.
- 1/π^2 terms: -1/π^2 + 1/π^2 = 0.
Thus, I = -1/6 - 2/π + 2 √3 / π, so p = -1/6, q = -2, r = 2.
Then, 2028 · p · q · r = 2028 · (-1/6) · (-2) · 2 = 2028 · (2/3) = 1352.