Area enclosed by parabola

Question

Area enclosed by parabola

The figure shows a graph of y= ax²+ bx + c. Suppose that the points (-h,y₀), (0,y₁), and (h,y₂) are on the graph. The area enclosed by the parabola, the x-axis, and the lines X = -h and x = h may be given by the formula below. Find the area enclosed by f(x)= x²+ 4x + 8, the x-axis, and the lines x = -4 and x = 4

area = h/3 (y₀+4y₁+y₂)

need help finding the area plzz

Answer 1

The area is given by the formula, so what you really need help with is how to determine y₀, y₁, and y₂

As (-h,y₀) is on the parabola, y₀ = f(-h) = a(-h)² +b(-h) +c = ah² - bh + c

As (0,y₁) is on the parabola, y₁ = f(0) = a(0)² +b(0) +c = c

As (h,y₂) is on the parabola, y₂ = f(h) = a(h)² +b(h) +c = ah² + bh + c

Then, A = h/3(y₀+ 4y₁ + y₂) = h/3(ah² - bh + c +4(c) + ah² + bh + c) = h/3(2ah² + 6c)= 2/3ah³ + 2hc

For f(x) = x² + 4x + 8, a = 1, b = 4, c = 8

h = 4

Thus, area = 2/3ah³ + 2hc = 2/3*1*4³ + 2 * 4 * 8 = 128/3 + 64 = 320/3 = 106 2/3