91Ó°ÊÓ

In mathematics, a partition of intervals is a way to break down a larger interval into smaller sub-intervals. This is crucial in calculating Riemann sums, which approximate the area under a curve. For example, given the partition \( \{0, 1, 2, 3\} \), we divide the interval from 0 to 3 into three smaller intervals: [0,1], [1,2], and [2,3].
These smaller intervals allow us to evaluate the function at specific points, helping us to estimate the area under the curve. In practical terms, partitions give us control over the granularity of our Riemann approximation. The more subintervals we have, the more accurate our approximation usually is, as we capture more variations of the function within the intervals.

The lower Riemann sum is a method for estimating the area under a curve by using the smallest value of a function within each sub-interval of a partition. In our example, we determine these minimum values across the intervals [0,1], [1,2], and [2,3].
To compute the lower Riemann sum:

• For [0,1], the minimum of \( f(x) = 1 \) since \( f(0) = 1 \).
• For [1,2], the minimum is \( f(2) = 5 \) instead of the other endpoint values.
• For [2,3], the minimum is \( f(3) = 1 \).

Using these minimum values, we multiply each by the interval's length (which is uniformly 1 in this case), then sum them: \[ 1 \times 1 + 5 \times 1 + 1 \times 1 = 7 \]. This lower sum provides a conservative estimate, often underestimating the true value of the integral.

The upper Riemann sum is another approximation method, but it uses the largest value of the function within each sub-interval of a partition. This approach tends to overestimate the area under the curve.
For our function over the partitions [0,1], [1,2], and [2,3]:

• For [0,1], the maximum is \( f(1) = 6 \).
• For [1,2], the maximum is again \( f(1) = 6 \).
• For [2,3], the maximum is \( f(2) = 5 \).

These maximum values are then multiplied by the interval width, similar to the lower sum: \[ 6 \times 1 + 6 \times 1 + 5 \times 1 = 17 \]. This upper sum provides an overestimate, creating a range that the actual integral value lies within.

The definite integral of a function over an interval gives the exact area under the curve from one point to another. In our exercise, this translates to finding \( \int_{0}^{3} f(x) \, dx \) for the function \( f(x) = 2x^3 - 9x^2 + 12x + 1 \).
Unlike Riemann sums, which are approximations, the definite integral calculates this precisely. In this context, evaluating the definite integral for our function actually gives us 12, the exact area under the curve from 0 to 3. This precise calculation should always fall between the lower and upper Riemann sums, providing a practical check: if calculated correctly, it confirms the methods and steps we've used in both the approximation processes.