Find the area bounded between the curves `y=x^(2), y=sqrt(x)`

Answer

To find the area bounded between the curves \( y = x^2 \) and \( y = \sqrt{x} \), we will follow these steps:

### Step 1: Find the Points of Intersection
To find the area between the curves, we first need to determine the points where the curves intersect. We set the equations equal to each other:

\[ x^2 = \sqrt{x} \]

Square both sides to eliminate the square root:

\[ (x^2)^2 = (\sqrt{x})^2 \implies x^4 = x \]

Rearranging gives us:

\[ x^4 - x = 0 \]

Factoring out \( x \):

\[ x(x^3 - 1) = 0 \]

This gives us the solutions:

\[ x = 0 \quad \text{or} \quad x^3 - 1 = 0 \implies x = 1 \]

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 \( x = 1 \) is given by the integral of the upper curve minus the lower curve. Here, \( y = \sqrt{x} \) is the upper curve and \( y = x^2 \) is the lower curve:

\[ A = \int_{0}^{1} (\sqrt{x} - x^2) \, dx \]

### Step 3: Calculate the Integral
Now we will compute the integral:

\[ A = \int_{0}^{1} \sqrt{x} \, dx - \int_{0}^{1} x^2 \, dx \]

Calculating each integral separately:

1. For \( \int \sqrt{x} \, dx \):

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

Evaluating from 0 to 1:

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

2. For \( \int x^2 \, dx \):

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

Evaluating from 0 to 1:

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

### Step 4: Combine the Results
Now we combine the results of the two integrals:

\[ A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

Thus, the area bounded between the curves \( y = x^2 \) and \( y = \sqrt{x} \) is:

\[ \boxed{\frac{1}{3}} \text{ square units} \]