[THOUGHT] The integral $I_n$ is given by $\frac{5}{3} (2^{n-1})^2 - (2^{n-1})$. We are given $I_n = \frac{1407374833221632}{3}$. Let $A = 2^{n-1}$. The equation is $\frac{5}{3} A^2 - A = \frac{1407374833221632}{3}$. Multiplying by 3, we get $5A^2 - 3A = 1407374833221632$. I will solve this quadratic equation for $A$ and then find $n = \log_2(A) + 1$. [/THOUGHT] [PYTHON] import math

target_value_numerator = 1407374833221632

Equation: 5A^2 - 3A - target_value_numerator = 0

Coefficients for quadratic formula ax^2 + bx + c = 0

a = 5 b = -3 c = -target_value_numerator

Calculate discriminant

D = b**2 - 4ac sqrt_D = math.isqrt(D)

if sqrt_D**2 == D: print("Discriminant is a perfect square.") A1 = ( -b + sqrt_D ) // (2a) A2 = ( -b - sqrt_D ) // (2a) print(f"A1 = {A1}") print(f"A2 = {A2}")

# Check if A1 is a power of 2
if A1 > 0 and (A1 & (A1 - 1) == 0):
    n = math.log2(A1) + 1
    print(f"Found n: {n}")

The integral of $f(x)$ from $a_n$ to $a_{n+1}$ is given by [ I_n = \int_{2^{n-1}-1}^{2^n-1} \left( \sqrt{2^{n-1}(x+1) - 2^{2n-2}} + 2^{n-1} - 1 \right) , dx. ] Let $A = 2^{n-1}$. Then the limits are $A-1$ to $2A-1$. The integrand is $\sqrt{A(x+1) - A^2} + A - 1$. Let $u = x+1$, so $du = dx$. The limits become $A$ to $2A$. [ I_n = \int_{A}^{2A} \left( \sqrt{Au - A^2} + A - 1 \right) , du = \int_{A}^{2A} \sqrt{A(u-A)} , du + \int_{A}^{2A} (A-1) , du. ] For the first part, let $v = u-A$, so $dv = du$. Limits $0$ to $A$. [ \int_{0}^{A} \sqrt{Av} , dv = \sqrt{A} \int_{0}^{A} v^{1/2} , dv = \sqrt{A} \left[ \frac{2}{3} v^{3/2} \right]_0^A = \sqrt{A} \cdot \frac{2}{3} A^{3/2} = \frac{2}{3} A^2. ] The second part is $(A-1)(2A - A) = A(A-1) = A^2 - A$. Thus, $I_n = \frac{2}{3} A^2 + A^2 - A = \frac{5}{3} A^2 - A$. We are given $I_n = \frac{1407374833221632}{3}$. [ \frac{5}{3} A^2 - A = \frac{1407374833221632}{3} \implies 5A^2 - 3A = 1407374833221632. ] Solving this quadratic equation for $A$ yields $A = 16777216$. Since $A = 2^{n-1}$, we have $2^{n-1} = 16777216 = 2^{24}$. Therefore, $n-1 = 24$, which implies $n = 25$.

\boxed{25}