### Related Formula Additivity property of definite integrals over split interval boundary points: int_a^c g(x) \, dx = int_a^b g(x) \, dx + int_b^c g(x) \, dx$\int_{a}^{c} g(x) \, dx = \int_{a}^{b} g(x) \, dx + \int_{b}^{c} g(x) \, dx$ ### Core Logic Separate the integration tracking flow into two component operations: I_1 = int_0^fracpi4 sinleft|4x - fracpi12 ight| \, dx$I_1 = \int_{0}^{\frac{pi}{4}} \sin\left|4x - \frac{pi}{12} ight| \, dx$ I_2 = int_0^fracpi4 [2sin x] \, dx$I_2 = \int_{0}^{\frac{pi}{4}} [2\sin x] \, dx$ ### Step 1: Solve Modulus Integral Term The argument changes sign inside absolute value bounds when 4x - fracpi12 = 0 implies x = fracpi48$4x - \frac{pi}{12} = 0 \implies x = \frac{pi}{48}$: I_1 = int_0^fracpi48 -sinleft(4x - fracpi12 ight) \, dx + int_fracpi48^fracpi4 sinleft(4x - fracpi12 ight) \, dx$I_1 = \int_{0}^{\frac{pi}{48}} -\sin\left(4x - \frac{pi}{12} ight) \, dx + \int_{\frac{pi}{48}}^{\frac{pi}{4}} \sin\left(4x - \frac{pi}{12} ight) \, dx$ = frac14 left[ cosleft(4x - fracpi12 ight) right]_0^fracpi48 - frac14 left[ cosleft(4x - fracpi12 ight) right]_fracpi48^fracpi4$= \frac{1}{4} \left[ \cos\left(4x - \frac{pi}{12} ight) \right]_{0}^{\frac{pi}{48}} - \frac{1}{4} \left[ \cos\left(4x - \frac{pi}{12} ight) \right]_{\frac{pi}{48}}^{\frac{pi}{4}}$ Evaluating across numerical boundaries gives: 24 cdot I_1 = 24left(frac12right) = 12$24 \cdot I_1 = 24\left(\frac{1}{2}\right) = 12$ ### Step 2: Solve Greatest Integer Component and Combine Analyze step limits inside greatest integer block [2sin x]$[2\sin x]$: For x in left[0, fracpi6right)$x \in \left[0, \frac{pi}{6}\right)$, 0 le 2sin x < 1 implies [2sin x] = 0$0 \le 2\sin x < 1 \implies [2\sin x] = 0$. For x in left[fracpi6, fracpi4right]$x \in \left[\frac{pi}{6}, \frac{pi}{4}\right]$, 1 le 2sin x < sqrt2 implies [2sin x] = 1$1 \le 2\sin x < \sqrt{2} \implies [2\sin x] = 1$. I_2 = int_0^fracpi6 0 \, dx + int_fracpi6^fracpi4 1 \, dx = fracpi4 - fracpi6 = fracpi12$I_2 = \int_{0}^{\frac{pi}{6}} 0 \, dx + \int_{\frac{pi}{6}}^{\frac{pi}{4}} 1 \, dx = \frac{pi}{4} - \frac{pi}{6} = \frac{pi}{12}$ Combine both sections aggregated by multiplier 24: textTotal = 12 + 24left(fracpi12 ight) = 2pi + 12$\text{Total} = 12 + 24\left(\frac{pi}{12} ight) = 2\pi + 12$ Comparing directly with expression statement parameters 2pi + alpha$2\pi + \alpha$ isolates response value: alpha = 12$\alpha = 12$ ### Pattern Recognition Modulus arguments and step functions require isolating inflection transition numbers directly to break integrations up neatly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration

### Related Formula Integration by Parts formula: int u cdot v \, dx = u int v \, dx - int left( u' int v \, dx right) dx$\int u \cdot v \, dx = u \int v \, dx - \int \left( u' \int v \, dx \right) dx$ ### Core Logic Given F(x) = xf(x)$F(x) = xf(x)$ and f(2) = 1 implies F(2) = 2f(2) = 2$f(2) = 1 \implies F(2) = 2f(2) = 2$. Let's apply Integration by Parts to the first given integral: int_0^2 x F'(x) dx = 6$\int_{0}^{2} x F'(x) dx = 6$ left[ xF(x) right]_0^2 - int_0^2 F(x) dx = 6$\left[ xF(x) \right]_0^2 - \int_{0}^{2} F(x) dx = 6$ 2F(2) - 0 - int_0^2 F(x) dx = 6$2F(2) - 0 - \int_{0}^{2} F(x) dx = 6$ 2(2) - int_0^2 F(x) dx = 6 implies int_0^2 F(x) dx = 4 - 6 = -2$2(2) - \int_{0}^{2} F(x) dx = 6 \implies \int_{0}^{2} F(x) dx = 4 - 6 = -2$ ### Step 1: Simplify Second Integral via Integration by Parts Now look at the second integral: int_0^2 x^2 F''(x) dx = 40$\int_{0}^{2} x^2 F''(x) dx = 40$ Applying Integration by parts (taking u = x^2$u = x^2$ and v = F''(x)$v = F''(x)$): left[ x^2 F'(x) right]_0^2 - int_0^2 2x F'(x) dx = 40$\left[ x^2 F'(x) \right]_0^2 - \int_{0}^{2} 2x F'(x) dx = 40$ 4 F'(2) - 0 - 2 int_0^2 x F'(x) dx = 40$4 F'(2) - 0 - 2 \int_{0}^{2} x F'(x) dx = 40$ We already know int_0^2 x F'(x) dx = 6$\int_{0}^{2} x F'(x) dx = 6$: 4 F'(2) - 2(6) = 40$4 F'(2) - 2(6) = 40$ 4 F'(2) - 12 = 40 implies 4 F'(2) = 52 implies F'(2) = 13$4 F'(2) - 12 = 40 \implies 4 F'(2) = 52 \implies F'(2) = 13$ ### Step 2: Sum the Values We need to find F'(2) + int_0^2 F(x) dx$F'(2) + \int_{0}^{2} F(x) dx$: 13 + (-2) = 11$13 + (-2) = 11$ ### Pattern Recognition Notice how the definition of f(x)$f(x)$ is mostly a distraction to find F(2)=2$F(2)=2$. The problem is fundamentally testing consecutive applications of integration by parts to reduction structures. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Definite Integration