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

tonys538 | Student

The area bounded by the curves `y=x^3` and `y=sqrt x` has to be determined.

The graph of the two curves gives an idea of what has to be done:

First we need to determine the points of intersection of the two curves. Equate `y = x^3 = sqrt x ` and solve for x.

`x^3 = sqrt x`

`x^3 - sqrt x = 0`

`sqrt x*(x^2.5 - 1) = 0`

`sqrt x = 0 => x = 0`

`x^2.5 - 1 = 0 => x^5 = 1 => x = 1`

The points of contact are at x = 0 and x = 1. This is the same as what is seen in the graph.

Now the area bounded by the curves is the integral:

`int_0^1 sqrt x - x^3 dx`

= `[x^(3/2)/(3/2) - x^4/4]_0^1`

= `2/3 - 0 - 1/4 + 0`

= `5/12`

The area bounded by the curves y = x^3 and `y = sqrt x` is equal to 5/12

giorgiana1976 | Student

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.