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

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To determine 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 intersection points:
$$
2x + 3 = x^2
$$
Rearrange the equation:
$$
x^2 - 2x - 3 = 0
$$
Solve the quadratic equation using the quadratic formula:
$$
x = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm \sqrt{16}}{2} = \frac{2 \pm 4}{2}
$$
Thus, the solutions are:
$$
x = 3 \quad \text{and} \quad x = -1
$$
So, the curves intersect at $$ x = -1 $$ and $$ x = 3 $$.

### 2. **Determine Which Function is on Top**

Choose a test point between $$ x = -1 $$ and $$ x = 3 $$, say $$ x = 0 $$:
$$
y_{\text{line}} = 2(0) + 3 = 3
$$
$$
y_{\text{parabola}} = (0)^2 = 0
$$
Since $$ y_{\text{line}} > y_{\text{parabola}} $$ at $$ x = 0 $$, the line $$ y = 2x + 3 $$ is above the parabola $$ y = x^2 $$ in this interval.

### 3. **Set Up the Integral**

The area $$ A $$ between the curves from $$ x = -1 $$ to $$ x = 3 $$ is:
$$
A = \int_{-1}^{3} \left( (2x + 3) - x^2 \right) \, dx
$$
Simplify the integrand:
$$
A = \int_{-1}^{3} (-x^2 + 2x + 3) \, dx
$$

### 4. **Evaluate the Integral**

Compute the indefinite integral:
$$
\int (-x^2 + 2x + 3) \, dx = -\frac{x^3}{3} + x^2 + 3x + C
$$
Evaluate from $$ x = -1 $$ to $$ x = 3 $$:
$$
A = \left[ -\frac{3^3}{3} + 3^2 + 3(3) \right] - \left[ -\frac{(-1)^3}{3} + (-1)^2 + 3(-1) \right]
$$
$$
A = \left[ -9 + 9 + 9 \right] - \left[ \frac{1}{3} + 1 - 3 \right]
$$
$$
A = 9 - \left( -\frac{5}{3} \right) = 9 + \frac{5}{3} = \frac{32}{3}
$$

### **Final Answer**

The area enclosed between the curves is $$ \dfrac{32}{3} $$ square units.
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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