Problem Analysis #20

To solve the problem, we need to find the area enclosed by the curve $y = f(x)$, the line $y = 2^{1012}(x - 2026)$, and the vertical lines $x = 2026$ and $x = 2027$.

Step-by-Step Solution

1. Understand the function $f(x)$:

   • For $x \in [0, 2]$, $f(x) = x \sin(\pi x)$.
   • For all $x$, $f(x) = 2 f(x - 2)$.

   This second property tells us that $f(x)$ is a scaled version of $f(x-2)$. Specifically, $f(x)$ doubles every 2 units.

2. Determine $f(x)$ for $x \in [2026, 2028]$:

   • Since $2026 = 1003 \times 2$, we have $f(2026) = 2^{1003} f(0)$.
   • Since $f(0) = 0 \sin(0) = 0$, $f(2026) = 0$.
   • For $x \in [2026, 2028]$, we can write $x = 2026 + t$ where $t \in [0, 2]$. Then:

$$f(2026 + t) = 2^{1003} f(t) = 2^{1003} (t \sin(\pi t)) = 2^{1003} t \sin(\pi t)$$

3. Find the intersection of $f(x)$ and the line $y = 2^{1012}(x - 2026)$:
   • Set $f(2026 + t) = 2^{1012}(t)$:

$$2^{1003} t \sin(\pi t) = 2^{1012} t$$

• Simplify by dividing both sides by $t$ (assuming $t \neq 0$):

$$2^{1003} \sin(\pi t) = 2^{1012}$$

• Divide both sides by $2^{1003}$:

$$\sin(\pi t) = 2^{1009}$$

• Since $\sin(\pi t)$ must be in the range $[-1, 1]$, there is no real $t$ that satisfies this equation. Therefore, the line $y = 2^{1012}(x - 2026)$ does not intersect $f(x)$ within $x \in [2026, 2027]$.

4. Calculate the area under $f(x)$ from $x = 2026$ to $x = 2027$:
   • The function $f(x) = 2^{1003} t \sin(\pi t)$ for $x = 2026 + t$.
   • We need to integrate $f(x)$ from $x = 2026$ to $x = 2027$:

$$\int_{2026}^{2027} f(x) \, dx = \int_{0}^{1} 2^{1003} t \sin(\pi t) \, dt$$

• Using integration by parts, let $u = t$ and $dv = \sin(\pi t) \, dt$. Then $du = dt$ and $v = -\frac{1}{\pi} \cos(\pi t)$:

$$\int t \sin(\pi t) \, dt = -\frac{t}{\pi} \cos(\pi t) \bigg|_0^1 + \frac{1}{\pi} \int \cos(\pi t) \, dt$$

• Evaluate the boundary term:

$$-\frac{t}{\pi} \cos(\pi t) \bigg|_0^1 = -\frac{1}{\pi} \cos(\pi) - \left( -\frac{0}{\pi} \cos(0) \right) = \frac{1}{\pi} - 0 = \frac{1}{\pi}$$

• Integrate $\cos(\pi t)$:

$$\int \cos(\pi t) \, dt = \frac{1}{\pi} \sin(\pi t)$$

• Evaluate the second term:

$$\frac{1}{\pi} \int_0^1 \cos(\pi t) \, dt = \frac{1}{\pi} \left( \frac{1}{\pi} \sin(\pi t) \bigg|_0^1 \right) = \frac{1}{\pi^2} (\sin(\pi) - \sin(0)) = 0$$

• Combine the results:

$$\int_0^1 t \sin(\pi t) \, dt = \frac{1}{\pi}$$

• Multiply by $2^{1003}$:

$$\int_{2026}^{2027} f(x) \, dx = 2^{1003} \cdot \frac{1}{\pi} = \frac{2^{1003}}{\pi}$$

5. Calculate the area under the line $y = 2^{1012}(x - 2026)$ from $x = 2026$ to $x = 2027$:
   • The line is $y = 2^{1012}(x - 2026)$.
   • Integrate from $x = 2026$ to $x = 2027$:

$$\int_{2026}^{2027} 2^{1012}(x - 2026) \, dx = 2^{1012} \int_{2026}^{2027} (x - 2026) \, dx$$

• Substitute $t = x - 2026$:

$$2^{1012} \int_{0}^{1} t \, dt = 2^{1012} \left( \frac{t^2}{2} \bigg|_0^1 \right) = 2^{1012} \cdot \frac{1}{2} = 2^{1011}$$

6. Calculate the area between the curve and the line:
   • The area is the difference between the area under the line and the area under the curve:

$$2^{1011} - \frac{2^{1003}}{\pi}$$

• Factor out $2^{1003}$:

$$2^{1003} \left( 2^{8} - \frac{1}{\pi} \right) = 2^{1003} \left( 256 - \frac{1}{\pi} \right)$$

7. Express in the given form:
   • The given form is $2^{1012} \left( p + \frac{q}{\pi} + \frac{r\sqrt{3}}{\pi} \right)$.
   • Compare:

$$2^{1003} \left( 256 - \frac{1}{\pi} \right) = 2^{1012} \left( \frac{256}{2^{11}} - \frac{1}{2^{9}\pi} + 0 \cdot \frac{\sqrt{3}}{\pi} \right)$$

• Simplify:

$$2^{1003} \left( 256 - \frac{1}{\pi} \right) = 2^{1012} \left( \frac{1}{8} - \frac{1}{512\pi} + 0 \cdot \frac{\sqrt{3}}{\pi} \right)$$

• Identify $p, q, r$:

$$p = \frac{1}{8}, \quad q = -\frac{1}{512}, \quad r = 0$$

8. Calculate $2028 \cdot p \cdot q \cdot r$:
   • Since $r = 0$:

$$2028 \cdot p \cdot q \cdot r = 2028 \cdot \frac{1}{8} \cdot \left( -\frac{1}{512} \right) \cdot 0 = 0$$

Thus, the final answer is $\boxed{0}$.