Find the area of the region bounded by: the parabola `y=x^2` and the line `y = x`

Answer

To find the area of the region bounded by the parabola \( y = x^2 \) and the line \( y = x \), we will follow these steps:

### Step 1: Find the Points of Intersection
To find the points where the parabola and the line intersect, we set the equations equal to each other:

\[ x^2 = x \]

Rearranging gives:

\[ x^2 - x = 0 \]

Factoring out \( x \):

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

Thus, the solutions are:

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

Now, we find the corresponding \( y \)-values:

- For \( x = 0 \): \( y = 0^2 = 0 \) → point \( (0, 0) \)
- For \( x = 1 \): \( y = 1^2 = 1 \) → point \( (1, 1) \)

### Step 2: Set Up the Integral for Area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral of the upper function minus the lower function. Here, the line \( y = x \) is above the parabola \( y = x^2 \) in this interval.

Thus, the area \( A \) is given by:

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

### Step 3: Evaluate the Integral
Now we compute the integral:

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

Calculating each integral separately:

1. For \( \int x \, dx \):

\[ \int x \, dx = \frac{x^2}{2} \bigg|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2} \]

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

\[ \int x^2 \, dx = \frac{x^3}{3} \bigg|_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

Now substituting back into the area formula:

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

### Step 4: Simplify the Result
To combine the fractions, we find a common denominator (which is 6):

\[ A = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Final Result
Thus, the area of the region bounded by the parabola \( y = x^2 \) and the line \( y = x \) is:

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

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