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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?

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### 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.

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

Learn how to evaluate the limit of a Riemann sum, \( \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{n} \left[ \left( \frac{4i}{n} \right)^3 + 3 \right] \), by converting it to a definite integral. This calculus problem explores concepts in integration and limit evaluation, suitable for students in Grades 11-12 or college calculus courses.

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