[TH

[THOUGHT] We need to compute the integral $\int_{a_n}^{a_{n+1}} f(x) \, dx$ where $a_n = 2^{n-1} - 1$. So $a_n = 2^{n-1} - 1$ and $a_{n+1} = 2^{n} - 1$. The integral is from $2^{n-1} - 1$ to $2^{n} - 1$. According to the definition of $f(x)$, on the interval $2^{n-1} - 1 \le x < 2^{n} - 1$, we have $f(x) = \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1$. Note that at the right endpoint $x = 2^{n} - 1$, the formula for $f(x)$ changes because it belongs to the next interval. But the integral is over a closed interval $[a_n, a_{n+1}]$. However, the function is defined piecewise, and at the right endpoint, the formula for the next piece might be different. But since $f$ is continuous, the value at the endpoint matches the limit from the left. So we can integrate using the given formula over the half-open interval and include the endpoint because the integral over a single point doesn't matter. So we can integrate $f(x)$ from $x = a_n$ to $x = a_{n+1}$ using the expression given for that interval.

Let’s set $n$ as a variable. We need to compute:

Simplify the integrand. Let $A = 2^{n-1}$. Then $a_n = A - 1$, $a_{n+1} = 2A - 1$. The expression inside the square root: $A(x+1) - A^2 = A(x+1 - A)$. So $f(x) = \sqrt{A(x+1 - A)} + A - 1$.

Thus:

Make substitution: Let $u = x+1 - A$. Then when $x = A-1$, $u = (A-1)+1 - A = 0$. When $x = 2A-1$, $u = (2A-1)+1 - A = A$. Also $dx = du$. Then:

Compute:

Thus:

Recall $A = 2^{n-1}$. So:

We are given that $I_n = \frac{1407374833221632}{3}$. So we solve for $n$:

Multiply both sides by 3:

Let $t = 2^{n-1}$. Then $2^{2n-2} = t^2$. Equation becomes:

Solve quadratic: $5 t^2 - 3 t - 1407374833221632 = 0$.

We need to find integer $t$ that is a power of 2. Compute discriminant and solve. Use Python to compute exact values.

We'll compute using Python to find $t$ and then $n = \log_2(t) + 1$. Ensure $n$ is a natural number.

Let's code. [/THOUGHT] [PYTHON] import sympy as sp

Given constant

C = 1407374833221632

Solve 5t^2 - 3t - C = 0 for t

t = sp.symbols('t', positive=True, integer=True) eq = 5t**2 - 3t - C solutions = sp.solve(eq, t) print("Solutions for t:", solutions)

Evaluate solutions numerically

for sol in solutions: print("Solution value:", sol) print("As integer:", int(sol) if sol.is_Integer else sol.evalf()) # Check if it's a power of two if sol.is_Integer: val = int(sol) # Check if val is a power of two if val > 0 and (val & (val - 1)) == 0: print("It is a power of two.") # compute n: t = 2^{n-1} => n-1 = log2(t) n = val.bit_length() # because 2^{k} has bit_length k+1? Actually, 2^{m} has bit_length m+1. # For example, 2^0 = 1, bit_length=1 => m=0 => m = bit_length-1. # So n-1 = bit_length - 1 => n = bit_length. n = val.bit_length() print("n =", n) # Verify: compute I_n from formula A = 2**(n-1) I_n = (5/3) * A2 - A print("I_n computed:", I_n) print("I_n exact rational:", sp.Rational(5,3)*A2 - A) else: print("Not a power of two.") [/PYTHON]

[THOUGHT] The Python output shows that $t = 16777216 = 2^{24}$, so $n-1 = 24$ giving $n = 25$. The computed integral matches the given value. Therefore, the natural number $n$ is 25. [/THOUGHT] $\boxed{25}$