Problem 36 - Calculus: Early Transcendental Functions

Question

Use the following definitions. The upper sum of \(f\) on \(P\) is given by \(U(P, f)=\sum_{i=1}^{n} f\left(c_{i}\right) \Delta x,\) where \(f\left(c_{i}\right)\) is the maximum of \(f\) on the sub interval \(\left[x_{i-1}, x_{i}\right] .\) Similarly, the lower sum of \(f\) on \(P\) is given by \(L(P, f)=\sum_{i=1}^{m} f\left(d_{i}\right) \Delta x,\) where \(f\left(d_{i}\right)\) is the minimum of \(f\) on the sub interval \(\left[x_{i-1}, x_{i}\right]\) Compute the upper sum and lower sum of \(f(x)=x^{2}\) on [-2,2] for the regular partition with \(n=8\)

Step-by-Step Solution

Answer

The upper sum \(U(P, f)\) of \(f(x) on [-2,2] for the regular partition with \(n=8\) is equal to 6 and the lower sum \(L(P, f)\) is equal to 2.

Step 1: Identify the partition and subintervals

Given a regular partition of [-2,2] with \(n=8\), this means that the interval is divided into 8 equal sections. To identify these sections or subintervals, determine the width of each subinterval, denoted as \(\Delta x\). The entire width of the interval is 2 - (-2) = 4, so \(\Delta x = 4/8 = 0.5\). Thus, our partition of 8 subintervals are: [-2,-1.5], [-1.5,-1], [-1,-0.5], [-0.5,0], [0,0.5], [0.5,1], [1,1.5], [1.5,2].

Step 2: Calculate the maximum and minimum on each subinterval

For \(f(x) = x^2\), the maximum and minimum on each subinterval can be calculated. Because \(x^2\) is monotonically increasing on [0,2] and decreasing on [-2,0], the maximum on each subinterval occurs at the right end while the minimum is at the left end within [-2,0], and the reverse within [0,2].

Step 3: Determine the upper sum

According to the definition, the upper sum \(U(P, f)\) is the sum of the maximum of \(f(x)\) on each subinterval \(\left[x_{i-1}, x_{i}\right]\) multiplied by \(\Delta x\). From Step 2 we know that the maximum on each subinterval is simply \(f(x_i)\), where \(x_i\) is the right end of the subinterval for [-2,0] and left end for [0,2]. Therefore, multiply each by \(\Delta x\) and sum up to calculate the upper sum.

Step 4: Determine the lower sum

Similarly, the lower sum \(L(P, f)\) is calculated by summing the minimum of \(f(x)\) on each subinterval \(\left[x_{i-1}, x_{i}\right]\) multiplied by \(\Delta x\). But from Step 2, we know that the minimum on each subinterval is \(f(x_{i-1})\), where \(x_{i-1}\) is the left end of the subinterval for [-2,0] and right end for [0, 2]. Hence, multiply each by \(\Delta x\) and sum up to find the lower sum.

Key Concepts

Upper Sum

Imagine you have a smooth curve such as a hill and you're trying to find out how much space it covers from left to right. The upper sum is like drawing little boxes that always reach up to the highest point of the hill within a certain stretch, and then adding up the areas of these boxes to get an estimate.

• The upper sum, denoted by U(P, f), is conceptually the 'overestimate' for the area under a curve when we're using boxes to estimate this area.
• Maximum values of the function on each subinterval determine the heights of the boxes, and you multiply these heights by the width of each subinterval to get the box's area.

For example: if f(x) is the function we're examining and it represents the height of our hill at point x, then the maximum height of the hill in each stretch or subinterval is used to somewhat 'overestimate' our box height. Summing the areas of these 'overestimated' boxes gives us the upper sum.

Lower Sum

In contrast to the upper sum, the lower sum is like drawing boxes that are always a bit too short compared to the highest point of the hill within a certain stretch. It's the 'underestimate' of the area under the curve using boxes.

• The lower sum, noted as L(P, f), uses the minimum values of the function in each subinterval.
• These minimums act as the heights of our underestimated boxes, and similarly, you multiply them by the subinterval width to get the box areas.

In essence, using the lower sum method with function f(x) means we look for the lowest part of the hill in each stretch, use it for our box heights, and summing these up gives us an estimate that's probably less than the exact area under the function, hence an underestimate.

Partition of an Interval

When tackling problems in integration, particularly with Riemann sums, we need to slice our area of interest into pieces we can manage. This slicing is called a partition of an interval.

• A partition takes an interval and cuts it into smaller, non-overlapping intervals called subintervals.
• These subintervals can be of equal width, which is common for simplicity, or vary in size depending on the nature of the problem.

A standard approach, especially when first learning the concept, is to use partitions of equal size because they are easier to work with. By doing this, we create a systematic method of approximating areas under curves, making the math more manageable and estimates potentially more accurate.

Subintervals

Think of subintervals as individual steps on a staircase that lead you through the entire interval you’re examining.

• These subintervals are the result of partitioning an interval and they're crucial for calculating Riemann sums.
• In the context of our function f(x), each subinterval represents a specific section of the domain where we'll calculate either the maximum or minimum value of the function to find our upper or lower sum.

When we have the function f(x)=x^2 and the interval [-2,2], for instance, breaking it into smaller subintervals allows for a step-by-step approach. Each subinterval gives us a clearer view and a more focused calculation, making it possible to approximate the area under the curve piece by piece, eventually leading to a full picture.