Bounded Integrals

Question

Set up an integral that would find the area in the first quadrant bounded on all sides by x=1, y = 2x^2 and y = 3+5x. You don't have to evaluate it. Please don't show In a video because I cant see them.

Answer 1

If 2x² = 5x+3, then 2x² - 5x - 3 = 0. So, (2x + 1)(x-3) = 0. Thus, x = -1/2 or 3.

The graph of y = 3 + 5x lies above the graph of y = 2x² between x = 1 and x = 3.

Area = ∫_{(1 to 3)} [(3 + 5x) - 2x²]dx

Answer 2

2x² = 3 + 5x;; x = (5 ± √49)/4; x = 3 (in 1st quadrant) and x = - 1/2 (not in 1st quadrant).

Area A = ∫₁³(3 + 5x - 2x²)dx = (3x + 5x²/2 - 2x³/3)₁³ = (9 + 45/2 - 18) - (3 + 5/2 - 2/3) = 27/2 - 29/6 = 26/3