Find the area of the region bounded by the graphs of the given curves

Question

y=x^3/2 and y=x^2/32

Answer 1

y=x^3/2 and y=x^2/32

We know that these two functions intersect at (0, 0). And

(x^3/2.)¹⁶ = (x^2/32)¹⁶

x²⁴ = x implies that x = 0, 1, showing additionally that the functions intersect also at (1, 1).

Take the definite integral of (x^2/32 - x^3/2) on the bound [0, 1]. So, this yields F(1) - F(0), where

F(x) = (32/34)x^34/32 - (2/5)x^5/2, whence F(1) = 32/34 - 2/5 = 46/85, and F(0) = 0 - 0 = 0, whence our desired result is 46/85.