Chapter 5: Problem 14
Let \(L_{n}\) denote the left-endpoint sum using \(n\) subintervals and let \(R_{n}\) denote the corresponding right-endpoint sum. In the following exercises, compute the indicated left and right sums for the given functions on the indicated interval. $$L_{6} \text { for } f(x)=\frac{1}{x(x-1)} \text { on }[2,5]$$
Short Answer
Step by step solution
Determine the Width of Subintervals
Identify the Left-Endpoints
Evaluate the Function at Each Left-Endpoint
Compute the Left-Endpoint Sum
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Left-Endpoint Sums
- Let's say we have an interval \[a, b\] divided into \ n \ subintervals.
- The left-endpoint sum, denoted as \ L_n \, sums up the areas of rectangles. Each rectangle's height is the function value at the left endpoint of each subinterval.
- The width of each of these rectangles is \ \Delta x \, the length of each subinterval.
Right-Endpoint Sums Explained
- Here, the height of each rectangle is the function value at the right endpoint of each subinterval.
- This means that the right endpoint of a subinterval \[x_{i+1}\] is used to compute the height for the corresponding rectangle.
Details on Subintervals
- The main interval \[a, b\] is partitioned into smaller, equally spaced segments.
- The number of subintervals, \ n \, dictates how fine these partitions will be.
- The width \ \Delta x \ of each subinterval is computed as \[\Delta x = \frac{b-a}{n}\].