The problem provided involves calculating the Riemann sum for the function $f(x) = \frac{2x^2}{x^2 + 1}$ over a certain interval. The graph shows a shaded region, and we are asked to compute the Riemann sum using $n = 60$ subintervals, selecting the left endpoint for each subinterval.

To approach this:

1. Function Analysis: The function is $f(x) = \frac{2x^2}{x^2 + 1}$, and we need to compute its Riemann sum over the shaded region (which looks to be from $x = 3$ to $x = 7$ based on the graph).

2. Riemann Sum Setup:

   • We divide the interval $[3, 7]$ into $n = 60$ subintervals.
   • The width of each subinterval is $\Delta x = \frac{7 - 3}{60} = \frac{4}{60} = \frac{1}{15}$.
   • For the left Riemann sum, we evaluate the function at the left endpoints $x_i = 3 + i \cdot \Delta x$ where $i = 0, 1, 2, \dots, 59$.

3. Riemann Sum Expression: The left Riemann sum $S$ is given by: $S \approx \sum_{i=0}^{59} f(3 + i \Delta x) \Delta x$ where $f(x) = \frac{2x^2}{x^2 + 1}$ and $\Delta x = \frac{1}{15}$.

Let me calculate the approximate value of the sum for you.The approximate value of the left Riemann sum for $n = 60$ subintervals over the interval $[3, 7]$ is $7.6349$, rounded to four decimal places.

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

Here are some follow-up questions to deepen your understanding:

1. How does the function $f(x) = \frac{2x^2}{x^2 + 1}$ behave at different intervals?
2. What is the significance of choosing the left endpoint for Riemann sums?
3. How does increasing the number of subintervals $n$ affect the accuracy of the Riemann sum?
4. How can we compute the exact value of the area using integrals?
5. How would the result change if we chose the right or midpoints for the Riemann sum?

Tip: When calculating Riemann sums, more subintervals $n$ usually result in greater accuracy as the approximation better reflects the curve's shape.