Calculate the area enclosed between the lines $$ y = 2x + 3 $$ and $$ y = x^2 $$.

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To calculate the area enclosed between the lines $$ y = 2x + 3 $$ and $$ y = x^2 $$, follow these steps:

### 1. Find the Points of Intersection
Set the equations equal to each other to find the points where the graphs intersect:

$$
2x + 3 = x^2 \
x^2 - 2x - 3 = 0
$$

Solve the quadratic equation:

$$
x = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}
$$

This gives the solutions:

$$
x = 3 \quad \text{and} \quad x = -1
$$

So, the points of intersection are at $$ x = -1 $$ and $$ x = 3 $$.

### 2. Determine Which Function is on Top
To find which function is above the other between $$ x = -1 $$ and $$ x = 3 $$:

- Choose a test point, say $$ x = 0 $$:
  $$
  y_1 = 2(0) + 3 = 3 \
  y_2 = (0)^2 = 0
  $$
  Here, $$ y = 2x + 3 $$ is above $$ y = x^2 $$.

This indicates that $$ y = 2x + 3 $$ is the upper function in this interval.

### 3. Set Up the Integral
The area $$ A $$ between the two curves from $$ x = -1 $$ to $$ x = 3 $$ is:

$$
A = \int_{-1}^{3} \left( (2x + 3) - x^2 \right) dx
$$

### 4. Compute the Integral
First, find the antiderivative:

$$
\int (2x + 3 - x^2) dx = x^2 + 3x - \frac{x^3}{3}
$$

Evaluate from $$ x = -1 $$ to $$ x = 3 $$:

$$
A = \left[ 3^2 + 3(3) - \frac{3^3}{3} \right] - \left[ (-1)^2 + 3(-1) - \frac{(-1)^3}{3} \right] \
= \left[ 9 + 9 - 9 \right] - \left[ 1 - 3 + \frac{1}{3} \right] \
= 9 - \left( -2 + \frac{1}{3} \right) \
= 9 + 2 - \frac{1}{3} \
= 11 - \frac{1}{3} \
= \frac{32}{3}
$$

### Final Answer
The area enclosed between the curves is:

$$
\boxed{\dfrac{32}{3}}
$$
The area between the lines $$ y = 2x + 3 $$ and $$ y = x^2 $$ is $$ \frac{32}{3} $$ square units.

Calculate the area enclosed between the lines and .

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