To solve part (a) of this problem, which asks for the approximate area using two rectangles under the curve $f(x) = 5 + x^2$ over the interval $[-2, 2]$, we'll follow these steps:

Step-by-step process:

1. Divide the interval into two rectangles:
   The interval $[-2, 2]$ has a total width of 4 units (since $2 - (-2) = 4$). With two rectangles, each rectangle will have a width of: $\text{Width of each rectangle} = \frac{4}{2} = 2$

2. Use the left endpoints to calculate the height:
   The heights of the rectangles are determined by the function values at the left side of each subinterval. So, we need to evaluate $f(x)$ at $x = -2$ and $x = 0$.

   • For the first rectangle, the left endpoint is at $x = -2$, so: $f(-2) = 5 + (-2)^2 = 5 + 4 = 9$

   • For the second rectangle, the left endpoint is at $x = 0$, so: $f(0) = 5 + 0^2 = 5$

3. Calculate the area of each rectangle:
   The area of a rectangle is given by $\text{Area} = \text{height} \times \text{width}$.

   • For the first rectangle (with left endpoint $x = -2$), the area is: $\text{Area} = 9 \times 2 = 18$

   • For the second rectangle (with left endpoint $x = 0$), the area is: $\text{Area} = 5 \times 2 = 10$

4. Add the areas of the two rectangles:
   $\text{Total area} = 18 + 10 = 28$

Thus, the approximate area under the curve using two rectangles is 28 square units.

Would you like more details or have any further questions on this process?

Here are 5 related questions to expand on this topic:

1. How would the approximation change if we used four rectangles instead of two?
2. What is the exact area under the curve from $-2$ to $2$ using integration?
3. How does the left Riemann sum compare to the right Riemann sum for this function?
4. Can you explain how the midpoint Riemann sum would be used in this case?
5. How does the approximation improve as the number of rectangles increases?

Tip:

When approximating areas under a curve, increasing the number of rectangles generally leads to a more accurate result, especially as the width of each rectangle becomes smaller.