To solve this problem, let’s analyze and interpret each part:

Problem Breakdown

The problem is about determining the integral $\int_a^b f(x) \, dx$ from a limit expression. You are given: $\lim_{n \to \infty} \sum_{i=1}^n \frac{2i}{n^2} \left(2 + \frac{2i}{n}\right)$ This expression resembles a Riemann sum that approximates an integral of the form $\int_a^b f(x) \, dx$.

Part (a): Determining $\Delta x$, $x_i$, $a$, $b$, and $f(x)$

To express the sum as a Riemann sum, let's interpret each part of the sum:

1. Riemann Sum Structure: A Riemann sum for $\int_a^b f(x) \, dx$ with $n$ subintervals has the form: $\sum_{i=1}^n f(x_i) \Delta x$ where:

   • $\Delta x = \frac{b - a}{n}$,
   • $x_i = a + i \Delta x$,
   • $f(x_i)$ represents the function evaluated at $x_i$.

2. Identifying $\Delta x$: By comparing the sum, $\frac{2i}{n^2} \left(2 + \frac{2i}{n}\right),$ we can see that $\frac{2}{n}$ likely represents $\Delta x$. Hence, $\Delta x = \frac{2}{n}$.

3. Identifying $x_i$: Since $\Delta x = \frac{2}{n}$, we can rewrite $x_i$ as: $x_i = a + i \Delta x.$ By inspecting the expression $2 + \frac{2i}{n}$ in the sum, we observe that: $x_i = 2 + \frac{2i}{n}.$ This implies $a = 2$ and $b = 4$ (as $x_i$ goes from $2$ to $4$ when $i$ goes from $1$ to $n$).

4. Identifying $f(x)$: The function $f(x)$ should match the remaining part of the sum that isn’t part of $\Delta x$. This means: $f(x) = x.$

Part (a) Solution Summary

Thus, we have:

• $\Delta x = \frac{2}{n}$,
• $x_i = 2 + \frac{2i}{n}$,
• $a = 2$,
• $b = 4$,
• $f(x) = x$.

Part (b): Calculating the Value of the Limit

The expression given is: $\lim_{n \to \infty} \sum_{i=1}^n \frac{2i}{n^2} \left(2 + \frac{2i}{n}\right).$ We can rewrite this as a Riemann sum for the integral: $\int_2^4 x \, dx.$ To calculate this integral: $\int_2^4 x \, dx = \left[ \frac{x^2}{2} \right]_2^4 = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6.$

Final Answer

For part (a): $(\Delta x, x_i, a, b, f(x)) = \left(\frac{2}{n}, 2 + \frac{2i}{n}, 2, 4, x\right).$

For part (b): $\lim_{n \to \infty} \sum_{i=1}^n \frac{2i}{n^2} \left(2 + \frac{2i}{n}\right) = 6.$

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. What is the purpose of using Riemann sums in approximating definite integrals?
2. How do different values of $a$ and $b$ affect the integral's value?
3. How would this problem change if $f(x)$ were a different function, such as $f(x) = x^2$?
4. Why does the limit as $n \to \infty$ give the exact value of the integral?
5. How would you interpret the meaning of $\Delta x$ and $x_i$ in practical terms?

Tip: Riemann sums become closer to the true integral value as the number of intervals $n$ increases, as the approximation error decreases with finer partitions.