Area enclosed by parabola, PLEASE HELP!

Question

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+ 2x + 4, the x-axis, and the lines x = -4 and x = 4

area = h/3 (y₀+4y₁+y₂)

the area is?

Answer 1

f(x) = x²+ 2x + 4

at x = -h = -4:

f(-4) = (-4)²+ 2(-4) + 4 = 12

(-h,y₀) = (-4,12)

at x = 0:

f(0) = (0)²+ 2(0) + 4 = 4

(0,y₁) = (0,4)

at x = h = 4:

f(4) = (4)²+ 2(4) + 4 = 28

(h,y₂) = (4,28)

area = h/3 (y₀ + 4y₁ + y₂) = (4/3) (12 + 4(4) + 28) = (4/3) (56) = 224/3 = 74.667

Answer 2

f(x) = y = x^2 +2x +4

y= x^2 + 2x + 1 + 3

y-3 = (x+1)^2, which is an upward opening parabola with vertex (-1,3)

It has x-intercepts where x^2 +2x +4 = 0

It never intersects the x axis

line of symmetry is x= -1 = -b/2a

y-intercept is y=4

to find the area below the parabola between x=-4 and 4

integrate the quadratic and evaluate it between those 2 limits

integral of x^2 +2x +4 is

x^3/3 +x^2 +4x

at x=4 that's 64/3 + 16 + 16 = 64/3 + 96/3 = 160/3

at x=-4 that's -64/3 +16 -16 = -64/3

160/3 - -64/3 = 224/3 = 74 2/3