If the area of the region { (x, y) : 1 + x² ≤ y ≤ min{ x + 7, 11 − 3x } } is A, then 3A is equal to:

❓ Question:

 Find the area of the region

$$\{(x,y):1+x^2\le y\le \min \{x+7,11-3x\}\}\mathrm{.}$$

If this area is $A$, compute $3A$.

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🖼️ Question Image

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✍️ Short Solution

1. Find where the two lines cross each other.
   Compare $x+7$ and $11-3x$:
   $x+7\le 11-3x⟺4x\le 4⟺x\le 1.$ So on $(-\infty,1]$ the upper boundary is $x+7$; on $[1,\infty)$ the upper boundary is $11-3x$. Both meet at $x=1$ with value $8$.

2. Find intersection points of parabola with each line.

   • With $y=x+7$: solve $1+x^{2}=x+7 \Rightarrow x^{2}-x-6=0$ → $x=-2,\,3$.

   • With $y=11-3x$: solve $1+x^2=11-3x\Rightarrow x^2+3x-10=\mathrm{0\; \to }$$x=-5,\,2$.

3. Determine the x-range where the parabola lies below the relevant line.

   • For the branch where upper = $x+7$ (valid for $x\le1$), the parabola is below this line on the interval between the intersection roots $[-2,3]$. Intersected with $x\le1$ gives $[-2,1]$.

   • For the branch where upper = $11-3x$ (valid for $x\ge1$), the parabola is below that line on $[-5,2]$. Intersected with $x\ge1$ gives $[1,2]$.

   So the region exists for $x\in[-2,2]$, split at $x=1$.

4. Set up the area integrals.

   $$A={\int }_{-2}^1[(x+7)-(1+x^2)]dx+{\int }_1^2[(11-3x)-(1+x^2)]dx\mathrm{.}$$

5. Compute the integrals.

   First integral:

   $$\int_{-2}^{1}(-x^{2}+x+6)\,dx =\Big[-\tfrac{x^{3}}{3}+\tfrac{x^{2}}{2}+6x\Big]_{-2}^{1} =\frac{27}{2}.$$
   

   Second integral:

   $$\int_{1}^{2}(-x^{2}-3x+10)\,dx =\Big[-\tfrac{x^{3}}{3}-\tfrac{3x^{2}}{2}+10x\Big]_{1}^{2} =\frac{19}{6}.$$
   

   Add them:

   $$A=\frac{27}{2}+\frac{19}{6}=\frac{81+19}{6}=\frac{100}{6}=\frac{50}{3}.$$
   

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🖼️ Image Solution

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✅ Conclusion & Video Solution

The area of the region is $A=\dfrac{50}{3}$. Therefore

$$\boxed{3A = 50}.$$