Problem 35 - 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 [0,2] for the regular partition with \(n=4\)

Step-by-Step Solution

Answer

The upper sum of \(f(x) = x^2\) on [0,2] for the regular partition with \(n=4\) is 1.375, while the lower sum is 0.625.

Step 1: Calculate the Length of Each Subinterval

The length of each subinterval can be calculated by dividing the length of the full interval by the number of subintervals. Therefore, \(\Delta x = \frac{2 - 0}{4} = 0.5\).

Step 2: Compute the Upper Sum

To compute the upper sum, we need to find the maximum value of \(f(x)\) in each subinterval, and sum these maximums. As \(f(x) = x^2\) is an increasing function on [0,2], the maximums occur at \(x_i\), the right endpoint of each subinterval. Subintervals are [0,0.5], [0.5,1], [1,1.5] and [1.5,2], thus \(U(P,f) = \sum_{i=1}^{4} f(x_{i}) \Delta x = 0.5^3 + 1^3 + 1.5^3 + 2^3 = 1.375\).

Step 3: Compute the Lower Sum

To compute the lower sum, we need to find the minimum value of \(f(x)\) in each subinterval, and sum these minimums. As \(f(x) = x^2\) is an increasing function on [0,2], the minimums occur at \(x_{i-1}\), the left endpoint of each subinterval. Subintervals are [0,0.5], [0.5,1], [1,1.5] and [1.5,2], thus \(L(P,f) = \sum_{i=1}^{4} f(x_{i-1}) \Delta x = 0^3 + 0.5^3 + 1^3 + 1.5^3 = 0.625\).

Key Concepts

Upper Sum

The upper sum is a method used in calculus to approximate the area under a curve. It represents an overestimation, assuming the highest possible value of the function within each interval. To calculate the upper sum, you divide the total interval into smaller subintervals and find the function's maximum value in each.

For instance, if you're working with the function f(x) = x² over the interval [0,2], you could break this interval into smaller parts, say four equal sections, making n=4. The length of each subinterval, or Δx, is the total interval length divided by n. For each subinterval, you then identify the right endpoint because for this function, the curve slopes upwards, meaning the maximum value within each segment will be at the rightmost point. Multiply these maximums by the length of subintervals and sum them to get the upper sum.

• Length of subinterval: Δx = 0.5
• Function values at right endpoints: f(0.5), f(1), f(1.5), and f(2)
• Sum of maximums: Σ f(x_i)Δx

The upper sum provides an overshoot of the integral, but it's useful for understanding the upper bounds of an area calculation.

Lower Sum

In contrast to the upper sum, the lower sum is an underestimation of the area under a curve. This method calculates the smallest area by taking the function's minimum value within each partitioned interval.

Using the same function, f(x) = x², and interval [0,2], the lower sum necessitates partitioning the interval into the same number of subintervals. However, you identify the minimum value of the function at the left endpoints because the function's value is lowest there. Multiply these minimum values by the subinterval length and sum them up to find the lower sum.

• Lower sum subintervals: Same as above, with Δx = 0.5
• Function values at left endpoints: f(0), f(0.5), f(1), f(1.5)
• Sum of minimums: Σ f(x_i-1)Δx

The lower sum gives you a baseline for how small the area could be, offering a foundation for comparison against more precise calculations.

Partition of an Interval

A partition of an interval is a crucial element of Riemann sums that involves dividing a continuous interval into a finite number of non-overlapping subintervals. This technique lays the groundwork for approximating areas and integrals.

The idea is to create smaller segments within the overall interval by specifying points along the interval. Each subinterval helps in estimating the area beneath the curve within that small section. If you are dealing with interval [0,2], partitioning it into four equal parts might mean creating points at x=0.5, x=1, x=1.5, and x=2.

• Number of subintervals: In our case, it's n=4
• Partition points: 0, 0.5, 1, 1.5, 2

Each of these subintervals will have its own contribution to the total area under the curve. The quality of your integral approximation improves with the number of subintervals, hence why choosing the right partition is vital.

Integral Approximation

Integral approximation is the essence of Riemann sums, serving as a foundational method for estimating the value of integrals, which represent areas under curves or other geometric properties in calculus. The process consists of computing either upper or lower sums or even better, taking more sophisticated approaches like the trapezoidal rule or Simpson's rule.

To illustrate with the function f(x) = x² over the interval [0,2], the goal is to find the area under the curve. Using upper and lower sums, you get different estimates that 'sandwich' the true area: the more partitions you use, the closer your approximation typically gets to the actual integral.

• Upper sum estimate: Overestimation
• Lower sum estimate: Underestimation
• Usefulness: Great for understanding the bounding values of integrals

This method, while not yielding exact values unless the function and partition are particularly compliant, provides immensely useful insights into the behavior of functions over an interval and is vital in the study of calculus.