[THOUGHT] To solve the problem, we start by analyzing the integral of the piecewise function ( f(x) ) over the interval ([a_n, a_{n+1})). The function ( f(x) ) is defined as: [ f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1 \quad \text{for} \quad 2^{n-1} - 1 \le x < 2^n - 1. ] We need to compute the integral: [ \int_{a_n}^{a_{n+1}} f(x) , dx = \int_{2^{n-1} - 1}^{2^n - 1} f(x) , dx. ]

First, we perform a substitution to simplify the integral. Let ( u = x + 1 ). Then ( du = dx ), and the limits of integration change as follows:

• When ( x = 2^{n-1} - 1 ), ( u = 2^{n-1} ).
• When ( x = 2^n - 1 ), ( u = 2^n ).

Thus, the integral becomes: [ \int_{2^{n-1}}^{2^n} \left( \sqrt{2^{n-1} u - 2^{2n-2}} + 2^{n-1} - 1 \right) , du. ]

Next, we simplify the expression inside the square root: [ 2^{n-1} u - 2^{2n-2} = 2^{n-1} (u - 2^{n-1}). ]

Let ( v = u - 2^{n-1} ). Then ( dv = du ), and the limits of integration change as follows:

• When ( u = 2^{n-1} ), ( v = 0 ).
• When ( u = 2^n ), ( v = 2^{n-1} ).

The integral now becomes: [ \int_{0}^{2^{n-1}} \left( \sqrt{2^{n-1} v} + 2^{n-1} - 1 \right) , dv. ]

We split this into two integrals: [ \int_{0}^{2^{n-1}} \sqrt{2^{n-1} v} , dv + \int_{0}^{2^{n-1}} (2^{n-1} - 1) , dv. ]

First, we evaluate the integral of the square root term: [ \int_{0}^{2^{n-1}} \sqrt{2^{n-1} v} , dv = \sqrt{2^{n-1}} \int_{0}^{2^{n-1}} \sqrt{v} , dv. ]

The integral of ( \sqrt{v} ) is: [ \int \sqrt{v} , dv = \frac{2}{3} v^{3/2}. ]

Evaluating this from 0 to ( 2^{n-1} ): [ \sqrt{2^{n-1}} \left[ \frac{2}{3} v^{3/2} \right]_{0}^{2^{n-1}} = \sqrt{2^{n-1}} \cdot \frac{2}{3} \left( (2^{n-1})^{3/2} - 0 \right) = \sqrt{2^{n-1}} \cdot \frac{2}{3} \cdot 2^{3(n-1)/2} = \frac{2}{3} \cdot 2^{(n-1)/2} \cdot 2^{3(n-1)/2} = \frac{2}{3} \cdot 2^{2(n-1)} = \frac{2}{3} \cdot 4^{n-1}. ]

Next, we evaluate the integral of the constant term: [ \int_{0}^{2^{n-1}} (2^{n-1} - 1) , dv = (2^{n-1} - 1) \int_{0}^{2^{n-1}} 1 , dv = (2^{n-1} - 1) \cdot 2^{n-1} = 4^{n-1} - 2^{n-1}. ]

Combining both results, the total integral is: [ \frac{2}{3} \cdot 4^{n-1} + 4^{n-1} - 2^{n-1} = \left( \frac{2}{3} + 1 \right) 4^{n-1} - 2^{n-1} = \frac{5}{3} \cdot 4^{n-1} - 2^{n-1}. ]

We are given that this integral equals: [ \frac{1407374833221632}{3}. ]

Thus, we set up the equation: [ \frac{5}{3} \cdot 4^{n-1} - 2^{n-1} = \frac{1407374833221632}{3}. ]

Multiplying through by 3 to clear the denominator: [ 5 \cdot 4^{n-1} - 3 \cdot 2^{n-1} = 1407374833221632. ]

Let ( k = 2^{n-1} ). Then ( 4^{n-1} = k^2 ), and the equation becomes: [ 5k^2 - 3k = 1407374833221632. ]

We solve this quadratic equation for ( k ): [ 5k^2 - 3k - 1407374833221632 = 0. ]

Using the quadratic formula ( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) with ( a = 5 ), ( b = -3 ), and ( c = -1407374833221632 ): [ k = \frac{3 \pm \sqrt{9 + 4 \cdot 5 \cdot 1407374833221632}}{10}. ]

Calculating the discriminant: [ 9 + 4 \cdot 5 \cdot 1407374833221632 = 9 + 28147496664432640 = 28147496664432649. ]

Taking the square root of the discriminant: [ \sqrt{28147496664432649} = 16777216. ]

Thus, the positive root is: [ k = \frac{3 + 16777216}{10} = \frac{16777219}{10} = 1677721.6. ]

Since ( k ) must be a power of 2, we check ( k = 2^{24} = 16777216 ): [ 5 \cdot (2^{24})^2 - 3 \cdot 2^{24} = 5 \cdot 2^{48} - 3 \cdot 2^{24} = 5 \cdot 281474976710656 - 3 \cdot 16777216 = 1407374883553280 - 50331648 = 1407374833221632. ]

This matches the given value. Therefore, ( k = 2^{24} ), so ( n-1 = 24 ) and ( n = 25 ).

[PYTHON]

Verification of the solution

n = 25 k = 2 ** (n - 1) result = 5 * (k ** 2) - 3 * k print(result == 1407374833221632) [/PYTHON]

[THOUGHT] The Python code confirms that ( n = 25 ) satisfies the equation. Therefore, the natural number ( n ) is:

[ \boxed{25} ]

[THOUGHT] The Python output confirms that ( n = 25 ) satisfies the equation. Therefore, the natural number ( n ) is:

[ \boxed{25} ]