### Related Formula The area enclosed between upper bounding function y_textupper$y_{\text{upper}}$ and lower function y_textlower$y_{\text{lower}}$ is: textArea = int_a^b (y_textupper - y_textlower) \, dx$\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) \, dx$ ### Core Logic We need to find the intersection points of the curves to understand the min\x+7, 11-3x\$\min\{x+7, 11-3x\}$ boundary transition: 1) x+7 = 11-3x implies 4x = 4 implies x = 1$x+7 = 11-3x \implies 4x = 4 \implies x = 1$. Hence, the line switches behavior at x=1$x=1$. 2) Intersecting 1+x^2$1+x^2$ with x+7$x+7$: x^2 - x - 6 = 0 implies (x-3)(x+2) = 0 implies x = -2 quad textor quad x = 3$x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0 \implies x = -2 \quad \text{or} \quad x = 3$ 3) Intersecting 1+x^2$1+x^2$ with 11-3x$11-3x$: x^2 + 3x - 10 = 0 implies (x+5)(x-2) = 0 implies x = 2 quad textor quad x = -5$x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0 \implies x = 2 \quad \text{or} \quad x = -5$

### Step 1: Set up Integrals The transition points show that from x = -2$x = -2$ to 1$1$, the upper line is x+7$x+7$, and from x = 1$x = 1$ to 2$2$, the upper line is 11-3x$11-3x$. A = int_-2^1 ((x + 7) - (1 + x^2)) \, dx + int_1^2 ((11 - 3x) - (1 + x^2)) \, dx$A = \int_{-2}^{1} ((x + 7) - (1 + x^2)) \, dx + \int_{1}^{2} ((11 - 3x) - (1 + x^2)) \, dx$ A = int_-2^1 (x + 6 - x^2) \, dx + int_1^2 (10 - 3x - x^2) \, dx$A = \int_{-2}^{1} (x + 6 - x^2) \, dx + \int_{1}^{2} (10 - 3x - x^2) \, dx$ ### Step 2: Integration Evaluation Evaluating the first integral: left[ fracx^22 + 6x - fracx^33 right]_-2^1 = left(frac12 + 6 - frac13right) - left(2 - 12 + frac83right) = frac376 - left(-frac223right) = frac272$\left[ \frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-2}^{1} = \left(\frac{1}{2} + 6 - \frac{1}{3}\right) - \left(2 - 12 + \frac{8}{3}\right) = \frac{37}{6} - \left(-\frac{22}{3}\right) = \frac{27}{2}$ Evaluating the second integral: left[ 10x - frac3x^22 - fracx^33 right]_1^2 = left(20 - 6 - frac83right) - left(10 - frac32 - frac13right) = frac343 - frac496 = frac196$\left[ 10x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} = \left(20 - 6 - \frac{8}{3}\right) - \left(10 - \frac{3}{2} - \frac{1}{3}\right) = \frac{34}{3} - \frac{49}{6} = \frac{19}{6}$ Total Area A$A$: A = frac272 + frac196 = frac81 + 196 = frac1006 = frac503$A = \frac{27}{2} + \frac{19}{6} = \frac{81 + 19}{6} = \frac{100}{6} = \frac{50}{3}$ ### Step 3: Calculate 3A Multiplying the total area by 3: 3A = 3 cdot left(frac503right) = 50$3A = 3 \cdot \left(\frac{50}{3}\right) = 50$ ### Pattern Recognition When a boundary contains a min\\$\min\{\}$ or max\\$\max\{\}$ component, always solve for their internal intersection first to identify the exact splitting point of your definite integrals. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus