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:

$$\boxed{22}$$