Problem Analysis #22

[THOUGHT] First, we need to understand the function ( f(x) ) and the sequence ( {a_n} ). The function ( f(x) ) is defined piecewise on intervals of the form ( [2^{n-1}-1, 2^n-1) ). For ( n = 1, 2, 3, \ldots ), the function is given by: [ f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1. ]

The sequence ( {a_n} ) is defined as ( a_n = 2^{n-1} - 1 ). Therefore, the integral we need to evaluate is: [ \int_{a_n}^{a_{n+1}} f(x) , dx = \int_{2^{n-1}-1}^{2^n-1} f(x) , dx. ]

Substituting ( f(x) ) into the integral, we have: [ \int_{2^{n-1}-1}^{2^n-1} \left( \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1 \right) , dx. ]

We can split this integral into two parts: [ \int_{2^{n-1}-1}^{2^n-1} \sqrt{2^{n-1}(x+1) - 2^{2n-2}} , dx + \int_{2^{n-1}-1}^{2^n-1} (2^{n-1} - 1) , dx. ]

First, let's evaluate the second integral: [ \int_{2^{n-1}-1}^{2^n-1} (2^{n-1} - 1) , dx = (2^{n-1} - 1) \left[ x \right]_{2^{n-1}-1}^{2^n-1} = (2^{n-1} - 1) \left( (2^n - 1) - (2^{n-1} - 1) \right) = (2^{n-1} - 1) \cdot 2^{n-1} = 2^{2n-2} - 2^{n-1}. ]

Next, let's evaluate the first integral. We make a substitution: [ u = 2^{n-1}(x+1) - 2^{2n-2}, ] [ du = 2^{n-1} , dx, ] [ dx = \frac{du}{2^{n-1}}. ]

When ( x = 2^{n-1} - 1 ), [ u = 2^{n-1}(2^{n-1}) - 2^{2n-2} = 0. ]

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

The integral becomes: [ \int_{0}^{2^{2n-2}} \sqrt{u} \cdot \frac{du}{2^{n-1}} = \frac{1}{2^{n-1}} \int_{0}^{2^{2n-2}} u^{1/2} , du. ]

Integrating ( u^{1/2} ): [ \int u^{1/2} , du = \frac{2}{3} u^{3/2}. ]

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

Adding the two integrals together: [ \int_{2^{n-1}-1}^{2^n-1} f(x) , dx = \frac{2^{2n-1}}{3} + 2^{2n-2} - 2^{n-1}. ]

We need this to equal: [ \frac{1407374833221632}{3}. ]

First, simplify ( \frac{2^{2n-1}}{3} + 2^{2n-2} - 2^{n-1} ): [ \frac{2^{2n-1}}{3} + 2^{2n-2} - 2^{n-1} = \frac{2^{2n-1} + 3 \cdot 2^{2n-2} - 3 \cdot 2^{n-1}}{3} = \frac{2^{2n-1} + 3 \cdot 2^{2n-2} - 3 \cdot 2^{n-1}}{3} = \frac{2 \cdot 2^{2n-2} + 3 \cdot 2^{2n-2} - 3 \cdot 2^{n-1}}{3} = \frac{5 \cdot 2^{2n-2} - 3 \cdot 2^{n-1}}{3}. ]

So we have: [ \frac{5 \cdot 2^{2n-2} - 3 \cdot 2^{n-1}}{3} = \frac{1407374833221632}{3}. ]

Multiply both sides by 3: [ 5 \cdot 2^{2n-2} - 3 \cdot 2^{n-1} = 1407374833221632. ]

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

Rearrange to form a quadratic equation: [ 5y^2 - 3y - 1407374833221632 = 0. ]

We can solve this quadratic equation using the quadratic formula ( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 5 ), ( b = -3 ), and ( c = -1407374833221632 ): [ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 5 \cdot (-1407374833221632)}}{2 \cdot 5} = \frac{3 \pm \sqrt{9 + 28014993168865280}}{10} = \frac{3 \pm \sqrt{28014993168865289}}{10}. ]

Calculate the discriminant: [ \sqrt{28014993168865289} = 52428800. ]

Thus: [ y = \frac{3 \pm 52428800}{10}. ]

Since ( y ) must be positive: [ y = \frac{52428803}{10} = 5242880.3. ]

Since ( y = 2^{n-1} ) must be an integer, we take: [ y = 5242880. ]

Therefore: [ 2^{n-1} = 5242880. ]

Taking the logarithm base 2: [ n-1 = \log_2(5242880) = 21. ]

Thus: [ n = 22. ]

So the natural number ( n ) is: [ \boxed{22} ]