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 ≤ y ≤ min { x + 7 , 11 − 3 x } } . If this area is A A , compute 3 A 3A . 🖼️ Question Image ✍️ Short Solution Find where the two lines cross each other. Compare x + 7 x+7 and 11 − 3 x 11-3x : x + 7 ≤ 11 − 3 x ⟺ 4 x ≤ 4 ⟺ x ≤ 1. So on ( − ∞ , 1 ] (-\infty,1] the upper boundary is x + 7 x+7 ; on [ 1 , ∞ ) [1,\infty) the upper boundary is 11 − 3 x 11-3x . Both meet at x = 1 x=1 with value 8 8 . Find intersection points of parabola with each line. With y = x + 7 y=x+7 : solve 1 + x 2 = x + 7 ⇒ x 2 − x − 6 = 0 1+x^{2}=x+7 \Rightarrow x^{2}-x-6=0 → x = − 2 , 3 x=-2,\,3 . With y = 11 − 3 x y=11-3x : solve 1 + x 2 = 11 − 3 x ⇒ x 2 + 3 x − 10 = 0 → x = − 5 , 2 x=-5,\,2 . Determine the x-range where the parabola lies below the relevant line. For the branch where upper = x + 7 x+7 (valid for x ≤ 1 x\le1 ), the parabola is below this...