The problem asks to find the area $A$ of the region between the curves $y = 2x^2 + 6$ and $y = 3x + 1$ over the interval $[-2, 3]$.

To solve this:

Step 1: Set up the integral

The area between two curves is given by the integral of the difference between the top curve and the bottom curve over the given interval. Here, the parabola $y = 2x^2 + 6$ is the top curve, and the line $y = 3x + 1$ is the bottom curve in the region of interest.

Thus, the area $A$ can be expressed as:

$A = \int_{-2}^{3} [(2x^2 + 6) - (3x + 1)] \, dx$

Step 2: Simplify the integrand

Simplify the expression inside the integral:

$A = \int_{-2}^{3} (2x^2 + 6 - 3x - 1) \, dx$ $A = \int_{-2}^{3} (2x^2 - 3x + 5) \, dx$

Step 3: Integrate the function

Now integrate each term separately: $\int 2x^2 \, dx = \frac{2}{3}x^3$ $\int -3x \, dx = -\frac{3}{2}x^2$ $\int 5 \, dx = 5x$

Thus, the integral becomes: $A = \left[\frac{2}{3}x^3 - \frac{3}{2}x^2 + 5x \right]_{-2}^{3}$

Step 4: Evaluate the definite integral

Now plug in the limits of integration:

At $x = 3$: $\frac{2}{3}(3)^3 - \frac{3}{2}(3)^2 + 5(3) = \frac{2}{3}(27) - \frac{3}{2}(9) + 15 = 18 - 13.5 + 15 = 19.5$

At $x = -2$: $\frac{2}{3}(-2)^3 - \frac{3}{2}(-2)^2 + 5(-2) = \frac{2}{3}(-8) - \frac{3}{2}(4) - 10 = -\frac{16}{3} - 6 - 10 = -\frac{16}{3} - 16 = -\frac{64}{3}$

Step 5: Subtract to find the total area

Now subtract the results: $A = 19.5 - \left(-\frac{64}{3}\right)$ $A = 19.5 + \frac{64}{3} = \frac{58.5}{1} + \frac{64}{3} = \frac{175.5 + 64}{3} = \frac{239.5}{3}$

Thus, the area is: $A = \frac{239.5}{3} \approx 79.83 \, \text{units}^2$

Let me know if you'd like further details or clarification.

Related Questions:

1. How do we know which curve is on top or bottom for the integral?
2. What is the significance of using definite integrals to find areas between curves?
3. Can this problem be solved using numerical methods instead of symbolic integration?
4. How would the limits of integration change if the interval was different?
5. What if the curves intersect within the interval? How would the area calculation change?

Tip: Always check for intersections between the curves within the interval before integrating, as this may change the setup of the area calculation.