To find the area of the region $R$ bounded by the curve $y = x^2 + 2x - 3$, the lines $x = -3$, $x = 1$, and $y = 0$, we need to integrate $y = x^2 + 2x - 3$ from $x = -3$ to $x = 1$ along the $x$-axis.

Step-by-Step Solution

1. Set up the Integral:

   The area $A$ of the region $R$ is given by: $A = \int_{-3}^{1} (x^2 + 2x - 3) \, dx$

2. Integrate the Function:

   We will find the integral of $x^2 + 2x - 3$: $\int (x^2 + 2x - 3) \, dx = \frac{x^3}{3} + x^2 - 3x$

3. Evaluate the Integral from $x = -3$ to $x = 1$:

   Substitute $x = 1$ and $x = -3$ into the antiderivative and find the difference.

   • For $x = 1$: $\frac{1^3}{3} + 1^2 - 3 \cdot 1 = \frac{1}{3} + 1 - 3 = \frac{1}{3} + 1 - 3 = \frac{1}{3} - 2 = -\frac{5}{3}$

   • For $x = -3$: $\frac{(-3)^3}{3} + (-3)^2 - 3 \cdot (-3) = \frac{-27}{3} + 9 + 9 = -9 + 9 + 9 = 9$

   Now, subtract the value at $x = -3$ from the value at $x = 1$: $A = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}$

Final Answer

The area of the region $R$ is: $A = \frac{32}{3} \text{ square units}$

Would you like further details or have any questions about these steps?

Here are 5 related questions for further exploration:

1. How would the area change if the boundaries were different?
2. What if the curve was shifted vertically or horizontally?
3. How would we find the area between two different curves?
4. How does changing the limits of integration affect the result?
5. How do you interpret the integral in terms of area under the curve?

Tip: Always check if the curve crosses the $x$-axis within the interval, as it affects whether to integrate directly or split the integral at the roots.