Problem Analysis #22

[THOUGHT]
The problem involves finding the natural number ( n ) such that the integral of ( f(x) ) from ( a_n ) to ( a_{n+1} ) equals ( \frac{1407374833221632}{3} ). The function ( f(x) ) is defined piecewise, and the sequence ( a_n = 2^{n-1} - 1 ).

The integral simplifies to:
[ \int_{a_n}^{a_{n+1}} f(x) dx = \frac{5}{3} \cdot 2^{2n-2} - 2^{n-1} ]

Setting this equal to the given fraction:
[ \frac{5}{3} \cdot 2^{2n-2} - 2^{n-1} = \frac{1407374833221632}{3} ]

Multiplying both sides by 3 to clear the denominator:
[ 5 \cdot 2^{2n-2} - 3 \cdot 2^{n-1} = 1407374833221632 ]

Simplifying the exponents:
[ 5 \cdot \frac{2^{2n}}{4} - 3 \cdot \frac{2^n}{2} = 1407374833221632 ]
[ \frac{5}{4} \cdot 2^{2n} - \frac{3}{2} \cdot 2^n = 1407374833221632 ]

Multiplying through by 4:
[ 5 \cdot 2^{2n} - 6 \cdot 2^n = 5629499332886528 ]

Substituting ( y = 2^n ):
[ 5y^2 - 6y = 5629499332886528 ]

Solving the quadratic equation:
[ 5y^2 - 6y - 5629499332886528 = 0 ]

Using the quadratic formula:
[ y = \frac{6 \pm \sqrt{36 + 4 \cdot 5 \cdot 5629499332886528}}{10} = \frac{6 \pm \sqrt{112589986657730596}}{10} ]

The discriminant ( 112589986657730596 ) is a perfect square. Calculating the square root:
[ \sqrt{112589986657730596} = 33554432 ]

Thus:
[ y = \frac{6 + 33554432}{10} = \frac{33554438}{10} = 3355443.8 \quad \text{(not an integer)} ]
But ( y = 2^n ) must be an integer. Re-evaluating, the correct calculation for ( y ) should yield an integer. Testing ( y = 2^{25} = 33554432 ):
[ 5 \cdot (33554432)^2 - 6 \cdot 33554432 = 5629499332886528 ]
This matches the right-hand side, so ( y = 33554432 = 2^{25} ), giving ( n = 25 ).

Verifying the integral:
[ \int_{a_{25}}^{a_{26}} f(x) dx = \frac{5}{3} \cdot 2^{48} - 2^{24} = \frac{5}{3} \cdot 281474976710656 - 16777216 = \frac{1407374883553280}{3} - 16777216 = \frac{1407374833221632}{3} ]
This matches the given value.

Thus, the natural number ( n ) is 25.

\boxed{25}

[THOUGHT] We need to solve for n in the equation 5y^2 - 6y = 5629499332886528, where y = 2^n. We'll use Python to compute the discriminant and verify if the root is a power of two. [/THOUGHT] [PYTHON] import math

C = 5629499332886528 D = 36 + 20 * C rootD = math.isqrt(D) if rootD * rootD == D: y = (6 + rootD) / 10 if y == int(y): y_int = int(y) if y_int & (y_int - 1) == 0: n = y_int.bit_length() - 1 print(n) else: print("Not a power of two") else: print("y is not integer") else: print("D is not a perfect square") [/PYTHON]

\boxed{25}