To approximate the sum of the cubes of the first $n$ positive integers, we can use the definite integral. The idea is that summing cubes like $1^3 + 2^3 + \dots + n^3$ can be approximated by the integral of the function $f(x) = x^3$.

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1. Sum of cubes formula (exact):

The exact formula for the sum of the first $n$ cubes is:

$\sum_{k=1}^{n} k^3 = \left( \frac{n(n + 1)}{2} \right)^2$

However, if $n$ is large, calculating this sum can become cumbersome, and using integration offers a useful approximation.

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2. Integral approximation:

We treat the sum as an integral of the function $f(x) = x^3$ over the interval from 1 to $n$. The idea is to replace the discrete sum with a continuous integral:

$\int_1^n x^3 \, dx$

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3. Solving the integral:

First, we compute the indefinite integral:

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

Now, apply the limits from 1 to $n$:

[ \int_1^n x^3 , dx = \left[ \frac{n^4}{4} \right] - \left[ \frac{1^4}{4} \right] = \frac{n^4}{4} - \frac{1}{4} ]

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4. Approximate sum of cubes:

Thus, the integral gives us an approximation for the sum of cubes:

$\sum_{k=1}^{n} k^3 \approx \frac{n^4}{4} - \frac{1}{4}$

For large $n$, the $-\frac{1}{4}$ term becomes negligible, so:

$\sum_{k=1}^{n} k^3 \approx \frac{n^4}{4}$

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5. Comparison with the exact formula:

The exact formula $\left( \frac{n(n + 1)}{2} \right)^2$ grows asymptotically close to $\frac{n^4}{4}$ for large $n$, confirming that the integral approximation is useful for estimating the sum.

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Would you like me to explain how to interpret this approximation further?

Here are 5 questions you could explore next:

1. How does the integral approximation improve as $n$ becomes larger?
2. Can you derive a similar approximation for the sum of squares $\sum_{k=1}^n k^2$?
3. How does the relative error between the exact and approximate sums change with $n$?
4. How do Riemann sums relate to the integral method used here?
5. How would the approximation change if we used a continuous sum starting from 0 instead of 1?

Tip: Integral approximations are powerful for large sums, especially when the exact formulas are complex or when you need only an estimate!