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

Answer 1

y^2 = 8x and y = x+2 touch each other at (2,4)
(find their intersection)

so your region consists of a right-angled triangle in quadrant II + the region between the y-axis, y^2 = 8x and y = x+2
all we really need Calculus for is the part on the right
that area = ∫ (x+2 - √8 x^(1/2) ) dx from x = 0 to 2
= [ x^2 /2 + 2x - (2/3)√8 x^(3/2) ] from 0 to 2
= [ x^2/2 + 2x - (4/3)(x^(3/2) ]
= 2 + 4 - (2/3)√8 √8 - (0+0+0)
= 6 - 16/3 = 2/3

add this to the triangle of area (1/2)(2)(2) = 2

total area = 2 + 2/3 = 8/3