A line parallel to x-axis bisects the region which was bounded by y = x^2and y= 4 into two regions of equal area. Find the equation of line.

Answer 1

draw a rough sketch. The graph of the area is between an upward opening parabola with vertex at the origin (0,0), and a horizontal line, that intersects the parabola at two points (2,4) and (-2,4)

if you drew a line parallel to the x-axis half way between y=4 and y=0, you'd get y=2, but you need a slightly higher y value, just from looking at the sketch. As the top half has larger area than the bottom half. Maybe y = 2.5 or a little higher

whatever you calculate after using integrals, you should get that general range.

integrate x^2 from 0 to 2. evaluate x^3/3 at 2 = 8/3

area in question is 8-8/3 = 16/3

half above, half below is 8/3

Let z^2 be the y value of the line dividing the two areas of equal size

0 and z are the limits of integration for x^2 = x^3/3 = z^3/3

z^2(z) - z^3/3 = 2z/3 = 8/3

2z^3 = 8

z^3 = 4

z = cube root of 4 = about 1.5874

z^2 = (1.5874)^2 = about 2.52

y = 2.52 is the line parallel to the x axis that cuts the area in two

Answer 2

Half the area:

(1/2)∫₀²(4-x²)dx = (1/2)[4x-x³/3]₀² = (1/2)(8-8/3) = 8/3

The bottom half of the area:

∫₀^x(y-t²)dt = yx-x³/3 = 8/3

But y=x², so

x³-x³/3 = (3x³-x³/3 = 2x³/3 = 8/3

x = 4^1/3

y = x² = (4^1/3)² = 4^2/3 = 16^1/3≈2.52