Question

If the area of the region $ \{(x, y) : 1 + x^2 \leq y \leq \min(x + 7, 11 - 3x)\} $ is $ A $, then $ 3A $ is equal to:

• A. 50
• B. 49
• C. 46
• D. 47

Hint:

When solving for areas between curves, always ensure to simplify the integrand and check your limits carefully.

Answer 1

The Correct Option is A

Understanding the Problem: The region is defined between two curves:

Lower boundary
: \( y = 1 + x^2 \) (a parabola opening upwards with vertex at \((0,1)\)).

Upper boundary
: \( y = \min(x + 7, 11 - 3x) \) (the minimum of two linear functions).
Step 1: Find the Point of Intersection of the Upper Boundary Functions
The upper boundary is defined by the minimum of: \[\begin{align} y &= x + 7 \\y &= 11 - 3x \end{align}\] Find their intersection point: \[\begin{align} x + 7 &= 11 - 3x \\4x &= 4 \\x &= 1 \end{align}\] At \( x = 1 \): \[ y = 1 + 7 = 8 \] So, the intersection is at \((1, 8)\). 
Step 2: Define the Upper Boundary Piecewise
The upper boundary can be written as:

• For \( x<1 \), \( y = x + 7 \) is the minimum.
• For \( x>1 \), \( y = 11 - 3x \) is the minimum.

Step 3: Find the Points of Intersection Between Lower and Upper Boundaries
Find where \( y = 1 + x^2 \) intersects the upper boundary. 
Case 1: \( x<1 \) (Upper boundary is \( y = x + 7 \))
\[\begin{align} 1 + x^2 &= x + 7 x^2 - x - 6 &= 0 \\x &= \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2} \\\Rightarrow x &= 3 \text{ or } x = -2 \end{align}\] Only \( x = -2 \) is valid since \( x<1 \). 
Case 2: \( x>1 \) (Upper boundary is \( y = 11 - 3x \))
\[\begin{align} 1 + x^2 &= 11 - 3x x^2 + 3x - 10 &= 0 \\x &= \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm 7}{2} \\\Rightarrow x &= 2 \text{ or } x = -5 \end{align}\] Only \( x = 2 \) is valid since \( x>1 \). 
Step 4: Determine the Range of Integration
The points of intersection are at \( x = -2 \) and \( x = 2 \). The upper boundary changes at \( x = 1 \), so we split the integral:

• From \( x = -2 \) to \( x = 1 \) (upper boundary: \( y = x + 7 \)).
• From \( x = 1 \) to \( x = 2 \) (upper boundary: \( y = 11 - 3x \)).

\subsection{Step 5: Calculate the Area} Part 1: \( x \) from \(-2\) to \(1\)} \[\begin{align} A_1 &= \int_{-2}^{1} \left[(x + 7) - (1 + x^2)\right] \, dx &= \int_{-2}^{1} (x + 6 - x^2) \, dx \\&= \left[ \frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-2}^{1} \\&= \left( \frac{1}{2} + 6 - \frac{1}{3} \right) - \left( 2 - 12 + \frac{8}{3} \right) \\&= \left( \frac{37}{6} \right) - \left( -\frac{22}{3} \right) \\&= \frac{37}{6} + \frac{44}{6} = \frac{81}{6} = \frac{27}{2} \end{align} \]
Part 2: \( x \) from \(1\) to \(2\)
\[\begin{align} A_2 &= \int_{1}^{2} \left[(11 - 3x) - (1 + x^2)\right] \, dx &= \int_{1}^{2} (10 - 3x - x^2) \, dx \\&= \left[ 10x - \frac{3x^2}{2} - \frac{x^3}{3} \right]_{1}^{2} \\&= \left( 20 - 6 - \frac{8}{3} \right) - \left( 10 - \frac{3}{2} - \frac{1}{3} \right) \\&= \left( \frac{34}{3} \right) - \left( \frac{49}{6} \right) \\&= \frac{68}{6} - \frac{49}{6} = \frac{19}{6} \end{align} \]
Total Area \( A \)
\[\begin{align} A &= A_1 + A_2 = \frac{27}{2} + \frac{19}{6} = \frac{81}{6} + \frac{19}{6} = \frac{100}{6} = \frac{50}{3} \end{align} \]
Calculate \( 3A \)
\[\begin{align} 3A &= 3 \times \frac{50}{3} = 50 \end{align} \]
Conclusion
The correct answer is 1.