Find the number b such that the line y = b divides the region bounded by the curves y = x^2 and y = 4 into two regions with equal area. Give your answer correct to 3 decimal places.

Answer 1

Hi Marietta,

As you know y = x² is parabola and even function so the area between and y = 4 and y = x² is 2 times of integral from ∫(4 - x²) dx from 0 to 2

∫(4 - x²) dx = 4x -1/3x³= 4 *2 -1/3 * 2^3 = 8 - 8/3

half of this area is 4 - 4/3 for ∫(b - x²)dx bounded 0 to √b ( area between y = x² and y = b)

∫(b - x²)dx = bx - 1/3 x³ by replace 0 and √b ( x² = b, x =±√b, we choose √b)

we will b√b - 1/3 (√b)³ = b√b - 1/3 b√b = 2/3 b√b

2/3 b√b = 4 -4/3 both side times 3

2b√b = 8 both side divided by 2√

b√b = 4

b^{3/2 = 4 both side to the power 2/3}

b = 16^1/3= 2.520 rounded to three decimal

i hope it is useful.

Answer 2

The line y=b will certainly be somewhere in between the line y=4 and the bottom of the parabola, at y=0. Since this is an area problem, integrals are a natural way to approach it. If we tried to integrate along x, we would need to know the points of intersection of y=b and y=x^2, which we currently don't, so it makes more sense to integrate along y, where the limits will be from 0 to b and from b to 4. Furthermore, since the parabola and the line are both symmetric about the y-axis, we need only consider the right half. Since we want the areas to be equal, the relevant equation is

∫_y=0^y=b √y dy = ∫_y=b^y=4 √y dy. This becomes:

(2/3)y ^3/2 ¦₀^b = (2/3)y ^3/2 ¦_b⁴.

(2/3)b ^3/2 = (2/3)4 ^3/2 - (2/3)b ^3/2

(4/3)b ^3/2 = (2/3)8

b ^3/2 = 4

b = 4 ^2/3

b = 2.520, rounded to three decimal places.

Hope this helps!