Problem 13 - Calculus

Question

Let \(P\) be a regular partition of the interval [0,2] into four equal subintervals, and let \(f(x)=x^{2}-1 .\) Write out the Riemann sum for \(f\) on \(P,\) in which \(\bar{x}_{i}\) is the right end point of each subinterval of \(P, i=1,2,3,4\). Find the value of this Riemann sum and make a sketch.

Step-by-Step Solution

Answer

The Riemann sum is 1.75.

Step 1: Understand the Partition

The interval [0,2] is divided into four equal subintervals. Each subinterval will have a length of \( \Delta x = \frac{2-0}{4} = 0.5 \). The right endpoints for these subintervals are 0.5, 1, 1.5, and 2.

Step 2: Identify Function Values at Right Endpoints

The function \( f(x) = x^2 - 1 \) needs to be evaluated at the right endpoints. Calculate these values:- For \( x = 0.5 \), \( f(0.5) = (0.5)^2 - 1 = -0.75 \).- For \( x = 1 \), \( f(1) = 1^2 - 1 = 0 \).- For \( x = 1.5 \), \( f(1.5) = (1.5)^2 - 1 = 1.25 \).- For \( x = 2 \), \( f(2) = 2^2 - 1 = 3 \).

Step 3: Apply the Riemann Sum Formula

The Riemann sum \( R \) is given by the formula \( R = \sum_{i=1}^{4} f(\bar{x}_i) \Delta x \). Plug in the values obtained:\[ R = [f(0.5) + f(1) + f(1.5) + f(2)] \times 0.5 \]\[ R = [-0.75 + 0 + 1.25 + 3] \times 0.5 = 3.5 \times 0.5 \]

Step 4: Calculate the Total Riemann Sum

Compute the total sum:\[ R = 3.5 \times 0.5 = 1.75 \]

Step 5: Sketch the Graph

Draw the function \( f(x) = x^2 - 1 \) on the interval [0,2]. Indicate the partition points at 0.5, 1, 1.5, and 2 and visualize the sum as the area of rectangles under the curve at the right endpoints.

Key Concepts

Regular Partition

In the context of Riemann sums, a **regular partition** refers to dividing an interval into subintervals of equal length. This ensures consistency when calculating the sum. Given an interval \[a, b\], if divided into \(n\) equal parts, each subinterval will have a length of \(\Delta x = \frac{b-a}{n}\). For our example from [0, 2], the interval is divided into 4 parts, making each subinterval's length \(\Delta x = 0.5\).

Regular partitions simplify the process of calculating integrals by providing uniform subintervals where function values can be easily calculated at specific points. This uniformity makes subsequent calculations straightforward and minimizes errors in evaluations, especially when using numerical methods like Riemann sums.

• Uniform subinterval length simplifies calculations.
• Ensures efficient and accurate computation in numerical integration.
• Aids in clear visualization without inconsistent partition sizes.

Right Endpoints

In calculating a Riemann sum, the choice of **right endpoints** is one of several methods to approximate the value of an integral. This involves selecting the point at the far right of each subinterval to evaluate the function. In our case, the right endpoints for the interval [0, 2] partitioned into four subintervals are 0.5, 1.0, 1.5, and 2.0.

Using right endpoints tends to overestimate or underestimate the integral depending on the behavior of the function on the interval:

• If the function is increasing, the right endpoint method typically leads to an overestimate because the function's value is higher at the endpoint than the start of the subinterval.
• If the function is decreasing, it usually results in an underestimate since the function's value is lower at the endpoint than at the beginning of the subinterval.

Subintervals

The term **subintervals** refers to the smaller sections an interval is divided into for evaluating the Riemann sum. Each subinterval essentially serves as an individual section over which the function value is calculated, providing a piecewise estimate of the integral of the function on the entire interval.

For the exercise, the interval [0, 2] is divided into four subintervals, each with a length of 0.5. The key process involves evaluating the function at the chosen point of each subinterval (here, the right endpoint), after which these values are summed up, each multiplied by the subinterval's length (width).

• The calculation results in an approximation of the integral over the entire interval.
• More subintervals generally lead to more accurate approximations, as they provide finer granularity in capturing the function's behavior across its range.
• Subintervals facilitate breaking complex calculations into more manageable parts.