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} ]