To approximate the sum of the cubes of the first $n$ whole numbers, we can use integration. We’ll denote the sum of cubes of the first $n$ numbers as:

$S_n = 1^3 + 2^3 + 3^3 + \dots + n^3$

The exact formula for this sum is:

$S_n = \left( \frac{n(n + 1)}{2} \right)^2$

However, to approximate this sum using integration, we can replace the sum with an integral. When $n$ is large, we can approximate $k^3$ by the continuous function $f(x) = x^3$. Thus:

$S_n \approx \int_0^n x^3 \, dx$

Step-by-Step Solution

1. Set up the integral:

   $\int_0^n x^3 \, dx$

2. Integrate $x^3$:

   To integrate $x^3$, we get:

   $\int x^3 \, dx = \frac{x^4}{4}$

3. Evaluate the integral from $0$ to $n$:

   $\int_0^n x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^n = \frac{n^4}{4}$

Approximation

Therefore, for large $n$, the sum of the cubes of the first $n$ whole numbers can be approximated by:

$S_n \approx \frac{n^4}{4}$

This integral approximation provides a close estimate of the actual sum for large $n$, with some error that decreases as $n$ increases.

Would you like a more detailed explanation, or have any specific questions?

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Related Questions:

1. How does the error in this approximation change as $n$ increases?
2. Can we use this technique to approximate sums of higher powers, like $k^4$?
3. How would the approach change if we wanted to approximate the sum of squares instead of cubes?
4. What other types of functions can we approximate sums for using integration?
5. Why is this approximation accurate only for large $n$?

Tip: For sums of higher powers, integration can approximate values, but always check for exact formulas where possible, especially for finite sums.