To find the area of the region bounded by the curves $y = x^2 + 2x - 4$ and $y = 3x + 2$, we need to:

1. Determine the points of intersection of the curves.
2. Set up the integral to find the area between the curves over the interval defined by these intersection points.

Step 1: Find the Points of Intersection

To find the points where $y = x^2 + 2x - 4$ intersects $y = 3x + 2$, set the two equations equal to each other:

$x^2 + 2x - 4 = 3x + 2$

Rearrange the equation:

$x^2 + 2x - 4 - 3x - 2 = 0$

$x^2 - x - 6 = 0$

Now, factor the quadratic equation:

$(x - 3)(x + 2) = 0$

This gives the solutions:

$x = 3 \quad \text{and} \quad x = -2$

These are the $x$-coordinates of the points of intersection. Now, substitute these values back into either original equation to find the corresponding $y$-values.

For $x = 3$:

$y = 3(3) + 2 = 9 + 2 = 11$

For $x = -2$:

$y = 3(-2) + 2 = -6 + 2 = -4$

So, the points of intersection are $(3, 11)$ and $(-2, -4)$.

Step 2: Set Up the Integral for the Area

The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$ is given by:

$\text{Area} = \int_a^b |f(x) - g(x)| \, dx$

In this case, from $x = -2$ to $x = 3$, the line $y = 3x + 2$ lies above the parabola $y = x^2 + 2x - 4$. So, we can set up the integral as:

$\text{Area} = \int_{-2}^{3} \left( (3x + 2) - (x^2 + 2x - 4) \right) \, dx$

Simplify the integrand:

$= \int_{-2}^{3} \left( 3x + 2 - x^2 - 2x + 4 \right) \, dx$

$= \int_{-2}^{3} \left( -x^2 + x + 6 \right) \, dx$

Step 3: Integrate

Now, integrate term-by-term:

$\int_{-2}^{3} \left( -x^2 + x + 6 \right) \, dx = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 6x \right]_{-2}^{3}$

Now, evaluate at the bounds:

1. For $x = 3$:

   $-\frac{3^3}{3} + \frac{3^2}{2} + 6 \cdot 3 = -\frac{27}{3} + \frac{9}{2} + 18 = -9 + 4.5 + 18 = 13.5$

2. For $x = -2$:

   $-\frac{(-2)^3}{3} + \frac{(-2)^2}{2} + 6 \cdot (-2) = \frac{8}{3} + 2 - 12 = \frac{8}{3} + 2 - 12 = \frac{8}{3} - 10 = -\frac{22}{3}$

Now, subtract the two results:

$\text{Area} = 13.5 - \left( -\frac{22}{3} \right) = 13.5 + \frac{22}{3} = \frac{40.5 + 22}{3} = \frac{62.5}{3} = 20.83$

Final Answer:

The area of the region bounded by the curves is approximately:

$\boxed{20.83}$

Would you like further details or have any questions?

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1. How would the solution change if the curves were shifted vertically?
2. What if the parabola opened downward? How would that affect the setup?
3. Could you calculate the area without finding exact intersection points?
4. What if we wanted to rotate this region around the x-axis? How would we calculate the volume?
5. How does the intersection point method help visualize bounded regions?

Tip: Always sketch the region when solving area problems between curves; it helps identify which curve is on top over each interval.