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

The region is defined by the inequalities
$$1+x^{2}\;\le\;y\;\le\;\min\{x+7,\;11-3x\}$$

First compare the two straight lines to find which is smaller at a given $$x$$.
Set $$x+7 = 11-3x$$ to locate their intersection:
$$4x = 4 \;\Longrightarrow\; x = 1,\qquad y = 8$$

Therefore
Case 1: $$x \le 1 \;$$ gives $$x+7 \lt 11-3x$$, so the upper curve is $$y = x+7$$.
Case 2: $$x \ge 1 \;$$ gives $$x+7 \gt 11-3x$$, so the upper curve is $$y = 11-3x$$.

The lower curve is always $$y = 1+x^{2}$$. To obtain the $$x$$-range where the region exists, ensure $$1+x^{2} \le \text{(upper curve)}$$ in each case.

Case 1: $$x \le 1$$
Require $$1+x^{2} \le x+7$$ $$\Longrightarrow\; x^{2}-x-6 \le 0$$ $$\Longrightarrow\; (x-3)(x+2) \le 0$$ $$\Longrightarrow\; -2 \le x \le 3$$. Combining with $$x \le 1$$ gives $$-2 \le x \le 1$$.

Case 2: $$x \ge 1$$
Require $$1+x^{2} \le 11-3x$$ $$\Longrightarrow\; x^{2}+3x-10 \le 0$$ $$\Longrightarrow\; (x+5)(x-2) \le 0$$ $$\Longrightarrow\; -5 \le x \le 2$$. Combining with $$x \ge 1$$ gives $$1 \le x \le 2$$.

Hence the complete $$x$$-interval for the region is $$-2 \le x \le 2$$, divided at $$x = 1$$.

The area $$A$$ is the sum of two integrals.

For $$-2 \le x \le 1$$
Upper curve: $$x+7$$, lower curve: $$1+x^{2}$$.
Area part $$A_{1}$$:
$$A_{1} = \int_{-2}^{1} \big[(x+7) - (1+x^{2})\big]\;dx = \int_{-2}^{1} \big(-x^{2}+x+6\big)\;dx$$

Integrate term-by-term:
$$\int (-x^{2})dx = -\frac{x^{3}}{3},\qquad \int x\,dx = \frac{x^{2}}{2},\qquad \int 6\,dx = 6x$$

Evaluate from $$x=-2$$ to $$x=1$$:
At $$x=1$$: $$-\tfrac{1}{3} + \tfrac{1}{2} + 6 = \tfrac{37}{6}$$
At $$x=-2$$: $$\;\;\tfrac{8}{3} + 2 - 12 = -\tfrac{22}{3}$$
Thus
$$A_{1} = \frac{37}{6} - \Big(-\frac{22}{3}\Big) = \frac{37}{6} + \frac{44}{6} = \frac{81}{6} = \frac{27}{2}$$

For $$1 \le x \le 2$$
Upper curve: $$11-3x$$, lower curve: $$1+x^{2}$$.
Area part $$A_{2}$$:
$$A_{2} = \int_{1}^{2} \big[(11-3x) - (1+x^{2})\big]\;dx = \int_{1}^{2} \big(-x^{2}-3x+10\big)\;dx$$

Integrate term-by-term:
$$\int (-x^{2})dx = -\frac{x^{3}}{3},\qquad \int (-3x)dx = -\frac{3x^{2}}{2},\qquad \int 10\,dx = 10x$$

Evaluate from $$x=1$$ to $$x=2$$:
At $$x=2$$: $$-\tfrac{8}{3} - 6 + 20 = \tfrac{34}{3}$$
At $$x=1$$: $$-\tfrac{1}{3} - \tfrac{3}{2} + 10 = \tfrac{49}{6}$$
Thus
$$A_{2} = \frac{34}{3} - \frac{49}{6} = \frac{68}{6} - \frac{49}{6} = \frac{19}{6}$$

Total area:
$$A = A_{1} + A_{2} = \frac{27}{2} + \frac{19}{6} = \frac{81}{6} + \frac{19}{6} = \frac{100}{6} = \frac{50}{3}$$

The question asks for $$3A$$:
$$3A = 3 \times \frac{50}{3} = 50$$

Therefore, $$3A = 50$$, which corresponds to Option A.