Find the area of the shape bounded by the lines y = x ^ 2 + 1 and y = 3

1. Find the intersection points of the graphs of the functions y = x ^ 2 + 1 and y = 3:
x ^ 2 + 1 = 3;
x ^ 2 = 2;
x1 = √ 2;
x2 = – √ 2;
2. The area of the figure bounded by the lines is equal to:
S = ∫ (√ 2; – √ 2) (x ^ 2 – 2) dx = x ^ 3/3 – 2 * x = (2 * √ 2/3 – 2 * √ 2) – (- 2 * √ 2 / 3 + 2 * √ 2) = (2 * √ 2/3 – 6 √ 2/3) – (- 2 * √ 2/3 + 6 √ 2/3) = – 4 √ 2/3 – 4 √ 2 / 3 = – 8 √ 2/3 = 8 √ 2/3 (the area cannot be negative);
3. Answer: S = 8 √ 2/3.