What is the area cut off the parabola 4y = 3𝑥^2 by the line 2y = 3x + 12 ?

1 Answer

44.859\ \text{unit}^2

Explanation:

Setting y=3/4x^2 in the equation of straight line: 2y=3x+12

2(3/4x^2)=3x+12

x^2-2x-8=0

(x-4)(x+2)=0

x=4, -2

The corresponding values of y are

y=12, 3

hence the line: 2y=3x+12 intersects the parabola at two points (4, 12) & (-2, 3)

Now, area region enclosed by the parabola & line is given as

\int (y_1-y_2)\ dx

Setting y_1=3/2x+6 & y_2=3/4x^2 & using proper limits from x=-2 to x=4,

=\int_{-2}^4 (3/2x+6-3/4x^2)\ dx

=(3/4x^2+6x-1/x^3)_{-2}^4

=3/4(4^2-(-2)^2)+6(4-(-2))-(1/(4)^3-1/{(-2)^3})

=2871/64

=44.859\ \text{unit}^2