The diagram shows the curve y^2=2x-1 and the straight line 3y=2x-1. The curve and the straight line intersect at x=1/2 and x=5, where a is a constant.

Question

Find, showing all necessary working, the area of the shaded region.

correct answer: 9/4

Answer 1

y^2=3y

y^2-3y + (3/2)^2 = (3/2)^2

y- 3/2 = +/-3/2

y= 0 or 3

3y=2x-1

2x-1=0 or 2x-1=9

x= 1/2 or 5

intersection points are (1/2, 0), (5,3)

area between the line and the rightward opening parabolic curve = 9/4 = 2.25

A= integral of (sqr(2x-1)-2x/3+1/3)

evaluated from 1/2 to 5

It may be easier to work with the inverse

switch x and y, then the intersection points are (0, 1/2) and (3,5)

the parabola is upward opening with vertex (0,1/2)

y^2 = 2x-1 becomes x^2 =2y-1 or y = x^2/2 +1/2

the line 3y = 2x-1 becomes 3x=2y-1 or y =3x/2 +1/2

integrate x^2/2 +1/2 - 3x/2 -1/2

= x^3/6 -3x/4

evalate from x=0 to x=3

= 3^3/6 -3(3)/4

= 9/2-9/4

=9/4= 2.25 = 2 1/4