Problem 11. \( \int_{0}^{\frac{3}{2}}\lfloor x+\lfloor 2 x\rfloor\rfloor \mathrm{d} x \) Problem 12. \( \int_{2025}^{2024} \ln (\{x\}) \mathrm{d} x \)

We first consider ### Problem 11 We need to evaluate \[ \int_{0}^{\frac{3}{2}} \lfloor x + \lfloor 2x \rfloor \rfloor \, dx. \] The key is to split the integration interval at the points where the inner floor function \(\lfloor 2x \rfloor\) changes value. Since \[ \lfloor 2x \rfloor \] jumps when \(2x\) is an integer, we have breakpoints at \[ x = \frac{n}{2}, \quad n=0,1,2,3. \] Thus, we split the interval \(\left[0,\frac{3}{2}\right]\) into: - \(x \in \left[0, \frac{1}{2}\right)\) - \(x \in \left[\frac{1}{2}, 1\right)\) - \(x \in \left[1, \frac{3}{2}\right]\) #### For \( x \in \left[0, \frac{1}{2}\right) \): - We have \(2x \in [0,1)\), so \(\lfloor 2x \rfloor = 0\). - Then, the integrand becomes: \[ \lfloor x + \lfloor 2x \rfloor \rfloor = \lfloor x + 0 \rfloor = \lfloor x \rfloor. \] Since \(x \in [0,0.5)\), we have \(\lfloor x \rfloor = 0\). - The contribution is: \[ \int_{0}^{\frac{1}{2}} 0 \, dx = 0. \] #### For \( x \in \left[\frac{1}{2}, 1\right) \): - We have \(2x \in [1,2)\), so \(\lfloor 2x \rfloor = 1\). - Then, the integrand becomes: \[ \lfloor x + \lfloor 2x \rfloor \rfloor = \lfloor x + 1 \rfloor. \] Since \(x \in \left[\frac{1}{2}, 1\right)\), it follows that \(x+1 \in \left[1.5,2\right)\). Thus, \(\lfloor x+1 \rfloor = 1\). - The contribution is: \[ \int_{\frac{1}{2}}^{1} 1 \, dx = 1\left(1 - \frac{1}{2}\right) = \frac{1}{2}. \] #### For \( x \in \left[1, \frac{3}{2}\right] \): - We have \(2x \in [2,3]\) (the endpoint \(3\) at \(x=\frac{3}{2}\) does not affect the integral), so \(\lfloor 2x \rfloor = 2\) almost everywhere on this interval. - Then, the integrand becomes: \[ \lfloor x + \lfloor 2x \rfloor \rfloor = \lfloor x + 2 \rfloor. \] For \(x \in \left[1, \frac{3}{2}\right]\), we have \(x+2 \in [3, 3.5]\), hence \(\lfloor x+2 \rfloor = 3\). - The contribution is: \[ \int_{1}^{\frac{3}{2}} 3 \, dx = 3\left(\frac{3}{2} - 1\right) = 3\left(\frac{1}{2}\right) = \frac{3}{2}. \] #### Summing the contributions: \[ 0 + \frac{1}{2} + \frac{3}{2} = 2. \] Thus, the answer for Problem 11 is \(2\). --- ### Problem 12 We need to evaluate \[ \int_{2025}^{2024} \ln (\{x\}) \, dx, \] where \(\{x\}\) denotes the fractional part of \(x\). Observing that the limits of integration are in reverse order, we have: \[ \int_{2025}^{2024} \ln (\{x\}) \, dx = - \int_{2024}^{2025} \ln (\{x\}) \, dx. \] Since the fractional part function \(\{x\}\) is periodic with period 1, for any integer \(n\) we have: \[ \int_{n}^{n+1} \ln (\{x\}) \, dx = \int_{0}^{1} \ln t \, dt. \] The integral \(\int_{0}^{1} \ln t \, dt\) is computed as follows: \[ \int_0^1 \ln t \, dt = \left. \left( t \ln t - t \right) \right|_0^1. \] At \(t = 1\): \[ 1 \cdot \ln 1 - 1 = 0 - 1 = -1. \] At \(t=0\), using the limit \(t \ln t \to 0\) as \(t \to 0^+\), we get \(0\). Thus, \[ \int_0^1 \ln t \, dt = -1. \] Therefore, \[ \int_{2024}^{2025} \ln (\{x\}) \, dx = -1. \] Accounting for the reversed limits: \[ \int_{2025}^{2024} \ln (\{x\}) \, dx = -\left(-1\right) = 1. \] Thus, the answer for Problem 12 is \(1\). --- ### Final Answers - Problem 11: \(2\) - Problem 12: \(1\)