The area enclosed between the curves `y = x^(3)` and `y = sqrt(x)` is

Options

A. `(5)/(3)` sq units
B. `(5)/(4)` sq units
C. `(5)/(12)` sq unit
D. `(12)/(5)` sq units

Explanation

To find the area enclosed between the curves \( y = x^3 \) and \( y = \sqrt{x} \), we will follow these steps:

### Step 1: Find the Points of Intersection
To find the points of intersection of the curves, we set \( y = x^3 \) equal to \( y = \sqrt{x} \):

\[ x^3 = \sqrt{x} \]

Squaring both sides to eliminate the square root gives:

\[ x^6 = x \]

Rearranging this, we have:

\[ x^6 - x = 0 \]

Factoring out \( x \):

\[ x(x^5 - 1) = 0 \]

This gives us \( x = 0 \) and \( x^5 - 1 = 0 \), which leads to \( x = 1 \). Thus, the points of intersection are \( x = 0 \) and \( x = 1 \).

### Step 2: Determine the Area Between the Curves
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral:

\[ A = \int_{0}^{1} (\sqrt{x} - x^3) \, dx \]

### Step 3: Evaluate the Integral
We will compute the integral \( \int (\sqrt{x} - x^3) \, dx \):

1. The integral of \( \sqrt{x} \) is:

\[ \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]

2. The integral of \( x^3 \) is:

\[ \int x^3 \, dx = \frac{x^4}{4} \]

Putting it all together, we have:

\[ A = \left[ \frac{2}{3} x^{3/2} - \frac{x^4}{4} \right]_{0}^{1} \]

### Step 4: Substitute the Limits
Now we will evaluate the definite integral from \( 0 \) to \( 1 \):

\[ A = \left( \frac{2}{3} (1)^{3/2} - \frac{(1)^4}{4} \right) - \left( \frac{2}{3} (0)^{3/2} - \frac{(0)^4}{4} \right) \]

This simplifies to:

\[ A = \left( \frac{2}{3} - \frac{1}{4} \right) - (0 - 0) \]

### Step 5: Simplify the Expression
Now we need to simplify \( \frac{2}{3} - \frac{1}{4} \):

To do this, we find a common denominator, which is \( 12 \):

\[ \frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12} \]

Thus,

\[ A = \frac{8}{12} - \frac{3}{12} = \frac{5}{12} \]

### Final Result
The area enclosed between the curves \( y = x^3 \) and \( y = \sqrt{x} \) is:

\[ \boxed{\frac{5}{12}} \text{ square units} \]