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

Options

A. `(9)/(2)` sq. units
B. `(9)/(4)` sq. units
C. `(9)/(8)` sq. units
D. 9 sq. units

Answer

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

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

\[ x^2 = x \]

Rearranging gives:

\[ x^2 - x = 0 \]

Factoring out \( x \):

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

This gives us the solutions:

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

### Step 2: Set Up the Integral
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral of the top function minus the bottom 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 \( \int_{0}^{1} x \, dx \):

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

2. For \( \int_{0}^{1} x^2 \, dx \):

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

Putting it all together:

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

### Step 4: Simplify the Result
To subtract these fractions, we need a common denominator, which is 6:

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

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

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