Problem Analysis #20

To solve the problem, we need to find 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 $x = 2027$.

Step 1: Understanding the Function $f(x)$

The function $f(x)$ is defined as:

$$f(x) = x \sin(\pi x) \quad \text{for} \quad x \in [0, 2]$$

and

$$f(x) = 2 f(x-2) \quad \text{for all real } x.$$

For $x \in [2026, 2027]$, we can express $x$ as $x = 2026 + t$ where $t \in [0, 1]$. Since $2026 = 2 \times 1013$, we have:

$$f(x) = 2^{1013} f(t) = 2^{1013} t \sin(\pi t).$$

Step 2: Setting Up the Integral

The line given is:

$$y = 2^{1012}(x - 2026) = 2^{1012} t.$$

The area between the curve $y = f(x)$ and the line $y = 2^{1012} t$ from $x = 2026$ to $x = 2027$ (or $t = 0$ to $t = 1$) is:

$$\text{Area} = \int_{0}^{1} \left[ 2^{1013} t \sin(\pi t) - 2^{1012} t \right] \, dt.$$

Factor out $2^{1012} t$:

$$\text{Area} = 2^{1012} \int_{0}^{1} t \left( 2 \sin(\pi t) - 1 \right) \, dt.$$

Step 3: Evaluating the Integral

We split the integral into two parts:

$$\int_{0}^{1} t \left( 2 \sin(\pi t) - 1 \right) \, dt = 2 \int_{0}^{1} t \sin(\pi t) \, dt - \int_{0}^{1} t \, dt.$$

Integral 1: $\int_{0}^{1} t \sin(\pi t) \, dt$

Using integration by parts with $u = t$ and $dv = \sin(\pi t) \, dt$:

$$du = dt, \quad v = -\frac{1}{\pi} \cos(\pi t).$$

Thus,

$$\int_{0}^{1} t \sin(\pi t) \, dt = \left[ -\frac{t}{\pi} \cos(\pi t) \right]_{0}^{1} + \frac{1}{\pi} \int_{0}^{1} \cos(\pi t) \, dt.$$

Evaluating the boundary terms:

$$\left[ -\frac{t}{\pi} \cos(\pi t) \right]_{0}^{1} = -\frac{1}{\pi} \cos(\pi) + 0 = \frac{1}{\pi}.$$

The remaining integral:

$$\frac{1}{\pi} \int_{0}^{1} \cos(\pi t) \, dt = \frac{1}{\pi} \left[ \frac{1}{\pi} \sin(\pi t) \right]_{0}^{1} = 0.$$

So,

$$\int_{0}^{1} t \sin(\pi t) \, dt = \frac{1}{\pi}.$$

Integral 2: $\int_{0}^{1} t \, dt$

$$\int_{0}^{1} t \, dt = \left[ \frac{t^2}{2} \right]_{0}^{1} = \frac{1}{2}.$$

Combining the results:

$$\int_{0}^{1} t \left( 2 \sin(\pi t) - 1 \right) \, dt = 2 \cdot \frac{1}{\pi} - \frac{1}{2} = \frac{2}{\pi} - \frac{1}{2}.$$

Step 4: Final Area Calculation

$$\text{Area} = 2^{1012} \left( \frac{2}{\pi} - \frac{1}{2} \right).$$

Step 5: Identifying $p, q, r$

Comparing with the given form:

$$2^{1012} \left( p + \frac{q}{\pi} + \frac{r \sqrt{3}}{\pi} \right),$$

we identify:

$$p = -\frac{1}{2}, \quad q = 2, \quad r = 0.$$

Step 6: Calculating the Final Product

$$2028 \cdot p \cdot q \cdot r = 2028 \cdot \left( -\frac{1}{2} \right) \cdot 2 \cdot 0 = 0.$$

Thus, the final answer is:

$$\boxed{0}$$