The problem in the image asks to determine the function $f(x)$ and region whose area is equal to the given limit expression. The provided options for $f(x)$ include different functions, and the interval $0 \leq x \leq 7$ appears to be relevant.

The limit in the question is represented as:

$\lim_{n \to \infty} \sum_{i=1}^{7} \frac{49 - \frac{49}{12+x_i^2}}{n}$

This looks like a Riemann sum, which approximates the area under a curve by dividing the region into small sub-intervals. The function being summed over seems to be based on $\sqrt{12 + x^2}$, indicating that the integral will involve this function over a specific range.

From the choices provided:

• Option A: $f(x) = \frac{\sqrt{12 + x^2}}{4}$, for $0 \leq x \leq 7$
• Option B: No Answer is correct
• Option C: $f(x) = \sqrt{49 + x^2} - 7 \leq x \leq 0$
• Option D: $f(x) = \sqrt{12 + x^2}$, for $0 \leq x \leq 7$
• Option E: $f(x) = \sqrt{12 + x^2}, 0 \leq x \leq 7$

Approach:

1. Interpret the limit as a Riemann sum for the integral.
2. Check if the given function matches the Riemann sum's integrand, which is likely $\sqrt{12 + x^2}$.
3. Compare this with the provided options.

The function inside the sum seems closest to the form in Option D and E. Specifically, $f(x) = \sqrt{12 + x^2}$ for $0 \leq x \leq 7$ is a plausible match, and the integral likely computes the area under this curve.

Thus, the correct choice appears to be Option D or E, depending on the formatting of the intervals.

Would you like further details on the solution, or have any follow-up questions?

Related questions:

1. What is the process of converting a Riemann sum to a definite integral?
2. How do you compute the definite integral of $\sqrt{12 + x^2}$?
3. What is the importance of limits in defining Riemann sums?
4. How is the choice of $f(x)$ related to the integrand in the Riemann sum?
5. Can the area under different functions be computed similarly using Riemann sums?

Tip:

When dealing with limits and sums, always check if they represent a Riemann sum, as this can often be converted into an integral for easier calculation.