We are approximating the area under the curve $f(x) = 5 + x^2$ over the interval $[-2, 2]$, using rectangles whose height is determined by the function's value at the left side of each rectangle. This is a common problem for applying the Left Riemann Sum method to approximate integrals.

Let's break it down step by step.

(a) Using 2 Rectangles:

• The interval $[-2, 2]$ has a total length of $4$ units.
• Dividing this interval into 2 equal subintervals, the width ($\Delta x$) of each rectangle is: $\Delta x = \frac{4}{2} = 2$
• The heights of the rectangles are determined by evaluating the function at the left endpoints $x = -2$ and $x = 0$: $f(-2) = 5 + (-2)^2 = 5 + 4 = 9$ $f(0) = 5 + 0^2 = 5$
• The area using 2 rectangles is: $A = 2 \times f(-2) + 2 \times f(0) = 2 \times 9 + 2 \times 5 = 18 + 10 = 28 \text{ square units.}$

This confirms that the approximated area when using two rectangles is 28 square units.

(b) Using 4 Rectangles:

• The interval $[-2, 2]$ is divided into 4 equal subintervals, so the width of each rectangle is: $\Delta x = \frac{4}{4} = 1$
• The heights of the rectangles are evaluated at the left endpoints $x = -2, -1, 0, 1$: $f(-2) = 5 + (-2)^2 = 9$ $f(-1) = 5 + (-1)^2 = 5 + 1 = 6$ $f(0) = 5 + 0^2 = 5$ $f(1) = 5 + 1^2 = 5 + 1 = 6$
• The area using 4 rectangles is: $A = 1 \times f(-2) + 1 \times f(-1) + 1 \times f(0) + 1 \times f(1)$ $A = 1 \times 9 + 1 \times 6 + 1 \times 5 + 1 \times 6 = 9 + 6 + 5 + 6 = 26 \text{ square units.}$

Thus, the approximated area when using four rectangles is 26 square units.

(c) Using 40 Rectangles:

• Now, dividing the interval into 40 subintervals, the width of each rectangle is: $\Delta x = \frac{4}{40} = 0.1$
• To find the total approximated area using 40 rectangles, we would evaluate the function at 40 equally spaced points starting from $x = -2$ to just before $x = 2$, and sum up the areas of all rectangles. However, for this step, we would typically use a calculator or technology.

Using a calculator or graphing software (such as a Riemann sum calculator), the approximated area using 40 rectangles is approximately:

$A \approx 21.47 \text{ square units.}$

Conclusion:

• (a) Using 2 rectangles: $A = 28$ square units.
• (b) Using 4 rectangles: $A = 26$ square units.
• (c) Using 40 rectangles: $A \approx 21.47$ square units.

Would you like further details or clarifications?

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Here are 5 related questions for further exploration:

1. How does the accuracy of the approximation change as the number of rectangles increases?
2. How would the Right Riemann Sum differ in this case?
3. Can you use the Midpoint Riemann Sum to approximate the area under this curve?
4. What is the exact value of the area under the curve using integration?
5. How would the approximation change if you used trapezoids instead of rectangles?

Tip: As the number of rectangles increases, the approximation becomes more accurate and converges to the exact area under the curve.