To find the area of the region $R$ bounded by the graphs of the functions:

$f(x) = x^2, \quad g(x) = 3 - 2x, \quad \text{and the line} \, y = 0,$

we'll follow these steps:

Step 1: Determine the points of intersection

The points of intersection of $f(x) = x^2$ and $g(x) = 3 - 2x$ can be found by setting the equations equal to each other:

$x^2 = 3 - 2x.$

Rearranging this into a standard quadratic form:

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

Factoring the quadratic:

$(x + 3)(x - 1) = 0.$

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

Step 2: Set up the integral for the area

The area $A$ of the region is given by the integral of the difference between the two functions from $x = -3$ to $x = 1$:

$A = \int_{-3}^{1} [g(x) - f(x)] \, dx.$

Substituting $g(x) = 3 - 2x$ and $f(x) = x^2$:

$A = \int_{-3}^{1} \left[(3 - 2x) - x^2\right] \, dx.$

Step 3: Simplify and integrate

Simplify the integrand:

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

Now, compute the integral term by term:

$A = \left[ 3x - x^2 - \frac{x^3}{3} \right]_{-3}^{1}.$

Step 4: Evaluate the integral

First, evaluate at $x = 1$:

$3(1) - (1)^2 - \frac{(1)^3}{3} = 3 - 1 - \frac{1}{3} = \frac{5}{3}.$

Next, evaluate at $x = -3$:

$3(-3) - (-3)^2 - \frac{(-3)^3}{3} = -9 - 9 + \frac{27}{3} = -18 + 9 = -9.$

Step 5: Calculate the area

Now subtract the two results:

$A = \frac{5}{3} - (-9) = \frac{5}{3} + 9 = \frac{5}{3} + \frac{27}{3} = \frac{32}{3}.$

Thus, the area of the region $R$ is:

$\boxed{\frac{32}{3}}.$

Would you like more details on any of the steps?

Here are 5 related questions:

1. How do we find the intersection points for more complex functions?
2. What is the geometric meaning of subtracting two functions in an integral?
3. How does the area calculation change if we integrate the functions over a different range?
4. Could the area between curves ever be negative, and why?
5. What happens if the region is defined by more than two curves?

Tip: Always visualize the region when setting up integrals; it helps in setting up limits of integration correctly!