How to find the area given the following?

Question

Find the area bounded by the parabola y² = 8x, the line y = x + 2, and the x-axis.

Answer 1

y^2 = 8x

y=x+2

y=0

x+2 = sqr8x

x^2 +4x + 4 = 8x

x^2-4x +4 = 0

(x-2)^2 = 0

x= 2 (2,2) is where the line and parabola intersect in quadrant I. That is the only intersectiion. the parabola is tangent to the line at that point. both have slope=1 at the point

(2,2). It helps to graph the lines.

the line and the x axis intersect at (-2,0)

from x=-2 to x= 0, the area under the line above the x axis = 2

from x=0 to x=2 the area under the line above the x axis = 6

2+6 = 8. Subtract from 8 the area under the parabola

integrate y=sqr8x which equals 2sqr2(x^3/2)/(3/2)

evaluate at x=2

(4/3)sqr2(2)sqr2=16/3

8- 16/3 = 8/3 = 2 2/3 = the area between the 3 lines, y^2=8x, y=x+2 and y= 0