To solve this problem, we are tasked with estimating the area under the curve $f(x) = \frac{5}{2 + x^2}$ using a Riemann sum with left endpoints and four rectangles of equal width between $x = 1$ and $x = 3$.

(a) List the four endpoints:

Since we need four intervals, the width of each rectangle is:

$\Delta x = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$

Thus, the left endpoints are:

$x_0 = 1, \quad x_1 = 1.5, \quad x_2 = 2, \quad x_3 = 2.5$

These values will be used to find the function values.

(b) Find the value of $f(x)$ at these points:

For $f(x) = \frac{5}{2 + x^2}$, we substitute each of the left endpoints $x_0, x_1, x_2, x_3$:

1. $f(x_0) = f(1) = \frac{5}{2 + 1^2} = \frac{5}{3} \approx 1.667$
2. $f(x_1) = f(1.5) = \frac{5}{2 + (1.5)^2} = \frac{5}{2 + 2.25} = \frac{5}{4.25} \approx 1.176$
3. $f(x_2) = f(2) = \frac{5}{2 + 2^2} = \frac{5}{6} \approx 0.833$
4. $f(x_3) = f(2.5) = \frac{5}{2 + (2.5)^2} = \frac{5}{2 + 6.25} = \frac{5}{8.25} \approx 0.606$

(c) Compute the Riemann sum:

The left Riemann sum is given by:

$L_4 = \Delta x \left[ f(x_0) + f(x_1) + f(x_2) + f(x_3) \right]$

Substituting the values we calculated:

$L_4 = 0.5 \times (1.667 + 1.176 + 0.833 + 0.606) = 0.5 \times 4.282 = 2.141$

Thus, the value of the Riemann sum $L_4 \approx 2.141$.

Would you like more details or clarifications? Here are 5 related questions to deepen your understanding:

1. What is the difference between left and right Riemann sums?
2. How do you calculate the Riemann sum using midpoints?
3. Why does increasing the number of rectangles make the Riemann sum more accurate?
4. How can the trapezoidal rule provide a more accurate approximation of the area?
5. What is the geometric interpretation of a Riemann sum?

Tip: As the number of rectangles increases, the approximation of the area under the curve becomes more accurate, approaching the actual integral value.