Integrate X²[x³] From -10 To 10: A Definite Guide

Hey guys! Let's dive into a fascinating problem involving definite integration. We're tasked with evaluating the definite integral of x²[x³] from -10 to 10. The twist? We need to be extra careful because of the greatest integer function, denoted by the square brackets [ ]. This function, also known as the floor function, gives us the greatest integer less than or equal to the number inside. This makes things a little trickier than your standard polynomial integration.

Understanding the Challenge

The main hurdle here is the greatest integer function, [x³]. Unlike a continuous function, the greatest integer function jumps at integer values. This means we can't just apply the usual integration rules directly. We need to think about how the function behaves within different intervals. Specifically, we need to figure out how [x³] changes as x varies from -10 to 10. You might initially think about checking if the function is odd to simplify the integral, but the presence of the greatest integer function throws a wrench in that plan. We can't simply assume [x³] behaves the same way an odd function would because it's not continuous. To get a handle on this, let's consider what happens when x is an integer and when it's not.

If x is an integer, [x³] is simply the cube of that integer. For instance, if x = 2, then [x³] = [2³] = [8] = 8. But when x is not an integer, things get more interesting. Let's say x = 2.5. Then x³ = (2.5)³ = 15.625, and [x³] = [15.625] = 15. Notice the jump? This jump discontinuity is what makes this problem a bit of a puzzle. So, how do we tackle this? The key is to break the integral into smaller intervals where [x³] is constant.

To effectively deal with this, we need to identify the points where [x³] changes its value. This happens precisely when x³ is an integer. So, we need to find the values of x for which x³ is an integer between (-10)³ = -1000 and 10³ = 1000. That's a lot of integers to consider! But don't worry, we won't have to calculate every single one. We'll use symmetry and some clever observations to simplify things.

Breaking Down the Integral

The strategy here is to split the integral into subintervals where [x³] remains constant. This means we're looking for integer values of x³. Let's denote these integers as 'n'. So, we have x³ = n, which means x = n^(1/3). These values of x will be our break points for the integral. We'll have a series of integrals, each over an interval where [x³] is a constant integer.

The integral we're looking at is:

∫[-10, 10] x²[x³] dx

We need to break this down into subintervals like:

∫[n^(1/3), (n+1)^(1/3)] x²[x³] dx

Within each of these subintervals, [x³] will be equal to 'n', an integer. This simplifies the integral within that interval to:

∫[n^(1/3), (n+1)^(1/3)] x² * n dx = n ∫[n^(1/3), (n+1)^(1/3)] x² dx

Now we can easily integrate x² within each interval. The result will be (n/3) * [x³] evaluated from n^(1/3) to (n+1)^(1/3). This gives us:

(n/3) * (((n+1)^(1/3))³ - (n^(1/3))³) = (n/3) * ((n+1) - n) = n/3

So, the integral over each subinterval is simply n/3, where 'n' is the integer value of [x³] in that interval. Now, we need to sum up these individual integrals over all relevant integer values of 'n' between -1000 and 1000.

Leveraging Symmetry to Simplify the Sum

Before we start summing, let's think about symmetry. Notice that for every positive integer 'n', there's a corresponding negative integer '-n'. This suggests there might be some cancellation that can simplify our work. Let's consider the integrals for 'n' and '-n':

For n: ∫[n^(1/3), (n+1)^(1/3)] x²[x³] dx ≈ n/3

For -n: ∫[(-n)^(1/3), (-n+1)^(1/3)] x²[x³] dx = ∫[(-n)^(1/3), (1-n)^(1/3)] x²[-n] dx ≈ -n/3

Here, we're using the fact that [-x] = -[x] - 1 if x is not an integer. If we pair up the contributions from 'n' and '-n', they appear to cancel out. However, we need to be careful about the intervals and the endpoints. The key is to notice that when we sum over all integer values of 'n' from -1000 to 999, most of the terms will indeed cancel out due to this symmetry.

Let's write out the sum we need to compute:

∑[n=-1000 to 999] n/3

This is a sum of integers from -1000 to 999, each divided by 3. If we pair the terms, we have:

(-1000/3 + 999/3) + (-999/3 + 998/3) + ... + (-1/3 + 0/3)

Most of these terms will cancel out, leaving us with only the -1000/3 term. This is because the sum of integers from -999 to 999 is zero, and we're left with the extra -1000 term.

So, the sum becomes:

-1000/3

Final Calculation and Considerations

We've done most of the heavy lifting. We broke down the integral, calculated the integral over subintervals, and used symmetry to simplify the summation. Now, let's put it all together. We found that the sum of the integrals over the subintervals is approximately -1000/3. But remember, this is just an approximation. We need to consider the exact limits of integration and the behavior of the greatest integer function near the endpoints.

The main thing we need to consider is the interval where x³ is close to -1000. When x = -10, x³ = -1000. So, we have the interval [-10, (-999)^(1/3)]. In this interval, [x³] = -1000. The contribution of this interval to the integral is:

∫[-10, (-999)^(1/3)] x²[-1000] dx = -1000 ∫[-10, (-999)^(1/3)] x² dx

Evaluating this integral gives us:

-1000/3 * [x³] from -10 to (-999)^(1/3) = -1000/3 * [(-999) - (-1000)] = -1000/3 * 1 = -1000/3

This confirms our earlier approximation. So, the definite integral of x²[x³] from -10 to 10 is indeed approximately -1000/3.

However, there's one more subtle point to consider. We've assumed that the sum of the integrals over the subintervals perfectly captures the total integral. But the greatest integer function has jump discontinuities, and we need to make sure we haven't missed any contributions due to these jumps. In particular, we need to examine the points where x³ transitions from one integer to the next.

To be absolutely rigorous, we would need to carefully analyze the behavior of the integral near each of these jump points. This involves taking limits and ensuring that the integral is well-defined. However, for the purpose of this discussion, we'll assume that these contributions are negligible due to the nature of the greatest integer function and the x² term, which smooths out the jumps to some extent.

Therefore, our final answer for the definite integral of x²[x³] from -10 to 10 is approximately -1000/3, or -333.33.

Conclusion

So, guys, we've successfully tackled this challenging integral involving the greatest integer function! We broke it down into smaller, manageable pieces, used symmetry to simplify the calculations, and carefully considered the behavior of the function near its discontinuities. This problem highlights the importance of understanding the properties of special functions like the greatest integer function and how they interact with integration. Remember, when dealing with such functions, always be mindful of the jump discontinuities and their potential impact on your results. Keep exploring and keep integrating!