Determine the area of the region bounded by y=x^3 and y=square root x.

First, we need to determine the intercepting points of the
given curves. These points will represent the necessary limits of
integration.

Since y = x^3 and y = `sqrt(x)` , we'll equate
and we'll get:

x^3 =
`sqrt(x)`

We'll raise to square both sides to remove the
square root:

x^6 = x

We'll
move the terms in x to the left:

x^6 - x =
0

We'll factorize by x:

x(x^5
- 1) = 0

x1 = 0 and x2 = 1

The
lower limit of integration is x = 0 and the upper limit is x =
1.

Since the graph of the function `sqrt(x)` is above the
graph of the function x^3, over the interval [0;1], we'll determine the area evaluating
the definite integral of the difference `sqrt(x)` -
x^3.

1                           1              
1

`int` `sqrt(x)` - x^3 dx = `int` `sqrt(x)` dx - `int`
x^3dx

0                        0               
0

1

`int` ` ` - x^3 dx =
[2x^(3/2)]/3 (0 -> 1) - x^4/4
(0->1)

0

We'll apply
Leibniz Newton
formula:

1

`int` ` ` - x^3 dx
= 2/3 -
1/4

0

1

`int`
` ` - x^3 dx = (8 -
3)/12

0

1

`int`
` ` - x^3 dx =
5/12

0

The area
of the region bounded by y = x^3 and y = `sqrt(x)` is A = 5/12 square
units.