Find the area, in square units, of the region determined by the intersections of y = x^1/2 and y = x^2 /512 ?

Question

I'm getting 64/3

Answer 1

Sqtr(x)=x^2/512, x1=0, x2=64;
I1=Integral [for x=0 until 64] of sqrt(x)*dx = 2x*sqrt(x)/3 = 16*(64/3)
I2=Integral [for x=0 until 64] of (x^2/512)*dx = x^3/(3*512) = 8*(64/3)
I1-I2=8*(64/3) – my answer!

Answer 2

Find the points where they intersect:

√x = x²/512
512√x = x²
262144x = x^4
0 = x^4 - 262144x = x(x³ - 262144) = x (x - 64) (something we don't care about)

So the bounds of the integral will be 0 and 64.


∫ x^½ - x²/512 dx = ⅔x^(3/2) - 1/(3*512)x³

Plugging in 64,

⅔(512) - 262144/(3*512) = ⅓(1024 - 512) = 512/3 = 170.667

If I didn't make any mistakes.

Answer 3

x^2/512=x^1/2
x^3/2=512
x^1/2=8
x=2rt2