Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b (f(x) - g(x)) \, dx$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$.
• Piecewise Functions: When dealing with $\min$ or $\max$ functions, identify the intervals where each function is the minimum or maximum, respectively.

Step-by-Step Solution

Step 1: Identify the Lower and Upper Boundaries

• What: Determine the functions that define the lower and upper boundaries of the region.
• Why: This is crucial for setting up the correct integral.
• The lower boundary is given by $y_L = 1 + x^2$.
• The upper boundary is given by $y_U = \min\{x+7, 11-3x\}$.

Step 2: Find the Intersection Point of the Two Linear Functions

• What: Find the $x$-value where $x+7 = 11-3x$.
• Why: This determines where the upper boundary switches from one linear function to the other.
• We have $x+7 = 11-3x$. Adding $3x$ to both sides and subtracting 7 from both sides gives $4x = 4$, so $x=1$.
• When $x=1$, $y = 1+7 = 8$. Thus, the two lines intersect at $(1, 8)$.

Step 3: Determine the Piecewise Definition of the Upper Boundary

• What: Express $y_U$ as a piecewise function.
• Why: The upper boundary is defined by the minimum of two functions, so we need to determine which function is smaller on different intervals.
• For $x < 1$, let $x=0$. Then $x+7 = 7$ and $11-3x = 11$. Since $7 < 11$, $y_U = x+7$ for $x < 1$.
• For $x > 1$, let $x=2$. Then $x+7 = 9$ and $11-3x = 5$. Since $5 < 9$, $y_U = 11-3x$ for $x > 1$.
• Therefore, $y_U(x) = \begin{cases} x+7 & \text{if } x \le 1 \\ 11-3x & \text{if } x > 1 \end{cases}$.

Step 4: Find the Intersection Points of the Parabola and the Linear Functions

• What: Find the $x$-values where $1+x^2 = x+7$ and $1+x^2 = 11-3x$.
• Why: These intersection points define the limits of integration.
• For $1+x^2 = x+7$, we have $x^2 - x - 6 = 0$, which factors as $(x-3)(x+2) = 0$. The solutions are $x=3$ and $x=-2$. Since we're considering the interval where $x \le 1$, we take $x=-2$.
• For $1+x^2 = 11-3x$, we have $x^2 + 3x - 10 = 0$, which factors as $(x+5)(x-2) = 0$. The solutions are $x=-5$ and $x=2$. Since we're considering the interval where $x > 1$, we take $x=2$.
• Thus, the region spans from $x=-2$ to $x=2$, and we need to split the integral at $x=1$.

Step 5: Set Up the Definite Integrals

• What: Write the integral for the area.
• Why: We need to integrate the difference between the upper and lower boundaries over the correct intervals.
• $A = \int_{-2}^1 [(x+7) - (1+x^2)] \, dx + \int_1^2 [(11-3x) - (1+x^2)] \, dx$
• $A = \int_{-2}^1 (-x^2 + x + 6) \, dx + \int_1^2 (-x^2 - 3x + 10) \, dx$

Step 6: Evaluate the First Integral

• What: Calculate $\int_{-2}^1 (-x^2 + x + 6) \, dx$.
• Why: This gives the area of the region from $x=-2$ to $x=1$.
• $\int_{-2}^1 (-x^2 + x + 6) \, dx = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 6x \right]_{-2}^1$
• $= \left( -\frac{1}{3} + \frac{1}{2} + 6 \right) - \left( -\frac{(-8)}{3} + \frac{4}{2} - 12 \right)$
• $= \left( -\frac{1}{3} + \frac{1}{2} + 6 \right) - \left( \frac{8}{3} + 2 - 12 \right) = -\frac{1}{3} + \frac{1}{2} + 6 - \frac{8}{3} - 2 + 12 = -\frac{9}{3} + \frac{1}{2} + 16 = -3 + \frac{1}{2} + 16 = 13 + \frac{1}{2} = \frac{27}{2}$

Step 7: Evaluate the Second Integral

• What: Calculate $\int_1^2 (-x^2 - 3x + 10) \, dx$.
• Why: This gives the area of the region from $x=1$ to $x=2$.
• $\int_1^2 (-x^2 - 3x + 10) \, dx = \left[ -\frac{x^3}{3} - \frac{3x^2}{2} + 10x \right]_1^2$
• $= \left( -\frac{8}{3} - \frac{12}{2} + 20 \right) - \left( -\frac{1}{3} - \frac{3}{2} + 10 \right)$
• $= \left( -\frac{8}{3} - 6 + 20 \right) - \left( -\frac{1}{3} - \frac{3}{2} + 10 \right) = -\frac{8}{3} + 14 + \frac{1}{3} + \frac{3}{2} - 10 = -\frac{7}{3} + 4 + \frac{3}{2} = \frac{-14 + 24 + 9}{6} = \frac{19}{6}$

Step 8: Calculate the Total Area

• What: Add the two integrals.
• Why: This gives the total area of the region.
• $A = \frac{27}{2} + \frac{19}{6} = \frac{81 + 19}{6} = \frac{100}{6} = \frac{50}{3}$

Step 9: Calculate 3A

• What: Find $3A$.
• Why: The problem asks for the value of $3A$.
• $3A = 3 \left( \frac{50}{3} \right) = 50$

Common Mistakes & Tips

• Incorrect Limits of Integration: Double-check that the limits of integration are the correct $x$-values where the curves intersect.
• Forgetting the Piecewise Nature: When dealing with $\min$ or $\max$, remember to split the integral into multiple integrals if necessary.
• Sign Errors: Be careful with signs when evaluating the definite integrals.

Summary

The area of the region is found by identifying the upper and lower boundary curves, finding their intersection points to determine the limits of integration, and integrating the difference between the upper and lower functions over the appropriate intervals. Because the upper boundary is defined as a minimum of two functions, we must express it as a piecewise function and split the integral accordingly. The total area $A$ is $\frac{50}{3}$, so $3A = 50$.

Final Answer

The final answer is \boxed{50}, which corresponds to option (A).