The problem you provided is:

$\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \left[ \left( \frac{4i}{n} \right)^3 + 3 \right]$

This expression is a Riemann sum, which suggests that it can be converted into an integral in the limit as $n \to \infty$.

Step-by-Step Solution

1. Identify the Riemann Sum Setup: The Riemann sum has the form: $\sum_{i=1}^n \frac{1}{n} f\left(x_i\right),$ where $x_i = \frac{4i}{n}$ and $f(x) = x^3 + 3$.

2. Convert to Integral Form: As $n \to \infty$, the Riemann sum approaches the integral of $f(x)$ over the interval that $x$ spans. Here, $x$ ranges from $\frac{4 \cdot 1}{n}$ to $\frac{4 \cdot n}{n}$, which is from 0 to 4.

   Therefore, the Riemann sum converts to: $\int_0^4 \left( x^3 + 3 \right) \, dx.$

3. Evaluate the Integral: We now integrate $f(x) = x^3 + 3$ over $[0, 4]$.

   $\int_0^4 \left( x^3 + 3 \right) \, dx = \int_0^4 x^3 \, dx + \int_0^4 3 \, dx.$

4. Calculate Each Integral:

   • For $\int_0^4 x^3 \, dx$: $\int x^3 \, dx = \frac{x^4}{4},$ so $\int_0^4 x^3 \, dx = \left. \frac{x^4}{4} \right|_0^4 = \frac{4^4}{4} - \frac{0^4}{4} = \frac{256}{4} = 64.$

   • For $\int_0^4 3 \, dx$: $\int_0^4 3 \, dx = 3x \Big|_0^4 = 3 \cdot 4 - 3 \cdot 0 = 12.$

5. Combine the Results:

   $\int_0^4 \left( x^3 + 3 \right) \, dx = 64 + 12 = 76.$

Final Answer

$\lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \left[ \left( \frac{4i}{n} \right)^3 + 3 \right] = 76.$

Would you like further clarification on any part of the solution?

Further Questions for Practice

1. How would the result change if the expression inside the sum was $\left( \frac{4i}{n} \right)^2 + 3$ instead of $\left( \frac{4i}{n} \right)^3 + 3$?
2. Can you identify another example of a Riemann sum that represents an integral over the interval $[1, 3]$?
3. How do you determine the limits of integration for a Riemann sum?
4. How would the Riemann sum change if the interval was split into subintervals of different lengths?
5. Why is the Riemann sum technique useful for approximating integrals?

Tip

When converting a Riemann sum to an integral, remember to carefully identify the interval and the function $f(x)$ being summed. This helps in setting up the correct integral.