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}).