The area between y = x and `y=x^(2), 0 le x le 2` is

Options

A. 1
B. `(2)/(3)`
C. `(1)/(3)`
D. `(4)/(3)`

Answer

To find the area between the curves \( y = x \) and \( y = x^2 \) from \( x = 0 \) to \( x = 2 \), we can follow these steps:

### Step 1: Find the Points of Intersection
We need to determine where the two curves intersect. Set \( y = x \) equal to \( y = x^2 \):

\[ x = x^2 \]

Rearranging gives:

\[ x^2 - x = 0 \]

Factoring out \( x \):

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

Thus, the solutions are:

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

### Step 2: Determine the Area Between the Curves
The area between the curves from \( x = 0 \) to \( x = 2 \) can be calculated by splitting the integral into two parts:

1. From \( x = 0 \) to \( x = 1 \), where \( y = x \) is above \( y = x^2 \).
2. From \( x = 1 \) to \( x = 2 \), where \( y = x^2 \) is above \( y = x \).

### Step 3: Set Up the Integrals
The area \( A \) can be expressed as:

\[ A = \int_0^1 (x - x^2) \, dx + \int_1^2 (x^2 - x) \, dx \]

### Step 4: Evaluate the First Integral
Calculate the first integral:

\[ \int_0^1 (x - x^2) \, dx = \int_0^1 x \, dx - \int_0^1 x^2 \, dx \]

Calculating each part:

\[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \]

\[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \]

Thus,

\[ \int_0^1 (x - x^2) \, dx = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

### Step 5: Evaluate the Second Integral
Now calculate the second integral:

\[ \int_1^2 (x^2 - x) \, dx = \int_1^2 x^2 \, dx - \int_1^2 x \, dx \]

Calculating each part:

\[ \int_1^2 x^2 \, dx = \left[ \frac{x^3}{3} \right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} \]

\[ \int_1^2 x \, dx = \left[ \frac{x^2}{2} \right]_1^2 = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]

Thus,

\[ \int_1^2 (x^2 - x) \, dx = \frac{7}{3} - \frac{3}{2} \]

Finding a common denominator (which is 6):

\[ \frac{7}{3} = \frac{14}{6}, \quad \frac{3}{2} = \frac{9}{6} \]

So,

\[ \int_1^2 (x^2 - x) \, dx = \frac{14}{6} - \frac{9}{6} = \frac{5}{6} \]

### Step 6: Combine the Areas
Now, combine both areas:

\[ A = \frac{1}{6} + \frac{5}{6} = 1 \]

### Final Answer
The area between the curves \( y = x \) and \( y = x^2 \) from \( x = 0 \) to \( x = 2 \) is:

\[ \boxed{1} \]