Find the area if the region bounded by `y=sqrtx" and " y=x`.

Explanation

To find the area of the region bounded by the curves \(y = \sqrt{x}\) and \(y = x\), we will follow these steps:

### Step 1: Find the points of intersection
We need to find the points where the two curves intersect. This is done by setting the equations equal to each other:

\[ \sqrt{x} = x \]

Squaring both sides gives:

\[ x = x^2 \]

Rearranging this equation, we get:

\[ x^2 - x = 0 \]

Factoring out \(x\):

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

This gives us the solutions:

\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 2: Determine the area between the curves
The area \(A\) between the curves from \(x = 0\) to \(x = 1\) can be calculated using the integral of the upper curve minus the lower curve. Here, \(y = \sqrt{x}\) is above \(y = x\) in this interval.

Thus, the area can be expressed as:

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

### Step 3: Evaluate the integral
Now we will evaluate the integral:

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

Calculating the first integral:

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

Evaluating from 0 to 1:

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

Now calculating the second integral:

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

Evaluating from 0 to 1:

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

### Step 4: Combine the results
Now substituting back into the area formula:

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

To combine these fractions, we need a common denominator, which is 6:

\[ A = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]

### Final Answer
Thus, the area of the region bounded by the curves \(y = \sqrt{x}\) and \(y = x\) is:

\[ \boxed{\frac{1}{6}} \text{ square units} \]