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Magazine Chapter 5: Problem 45
For the functions find a formula for the Riemann sum obtained by dividing the interval \([a, b]\) into \(n\) equal subintervals and using the right-hand endpoint for each \(c_{k} .\) Then take a limit of these sums as \(n \rightarrow \infty\) to calculate the area under the curve over \([a, b] .\) $$ f(x)=x^{2}+1 \text { over the interval }[0,3]. $$
Short Answer
Expert verified
The area under the curve is 12.
Step by step solution
01
Divide the Interval Into Subintervals
To begin, divide the interval \([0, 3]\) into \(n\) equal subintervals. The width of each subinterval, \(\Delta x\), is \(\Delta x = \frac{b-a}{n} = \frac{3-0}{n} = \frac{3}{n}\).
02
Define Right-Hand Endpoints
For a right-hand Riemann sum, use the right endpoint of each subinterval for evaluating the function. Let the \(k\)-th subinterval be \([x_{k-1}, x_{k}]\), where \(x_{k} = a + k\Delta x = 0 + k\left(\frac{3}{n}\right) = \frac{3k}{n}\).
03
Set Up the Riemann Sum
The Riemann sum \(R_n\) using right-hand endpoints is given by \(R_n = \sum_{k=1}^{n} f(x_{k})\Delta x\). Substituting \(f(x) = x^2 + 1\) and \(x_{k} = \frac{3k}{n}\), we have \[R_n = \sum_{k=1}^{n} \left(\left(\frac{3k}{n}\right)^2 + 1\right)\frac{3}{n}.\]
04
Simplify the Riemann Sum
Simplify the Riemann sum: \[R_n = \sum_{k=1}^{n} \left(\frac{9k^2}{n^2} + 1\right)\frac{3}{n} = \frac{3}{n} \sum_{k=1}^{n} \left(\frac{9k^2}{n^2} + 1\right).\] Distribute \(\frac{3}{n}\): \[R_n = \frac{27}{n^3} \sum_{k=1}^{n} k^2 + \frac{3}{n} \sum_{k=1}^{n} 1.\]
05
Calculate the Sums
Use the formula for the sum of squares \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\) and the sum of ones \(\sum_{k=1}^{n} 1 = n\): \[R_n = \frac{27}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} + \frac{3n}{n}.\] Simplify: \[R_n = \frac{27(n+1)(2n+1)}{6n^2} + 3.\]
06
Take the Limit as n Approaches Infinity
Now, take the limit as \(n \rightarrow \infty\): \[\lim_{n \to \infty} R_n = \lim_{n \to \infty} \left(\frac{27(n+1)(2n+1)}{6n^2} + 3\right).\] Simplify each term: Divide \(27(2+\frac{3}{n}+\frac{1}{n^2})\) by 6, which yields 9 as the leading term, and as \(n\to \infty\), the fraction part approaches 0. Thus, the limit is 12.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit of Sums
The concept of the limit of sums is central to understanding Riemann sums and their application to finding areas under curves. When calculating the area under a curve using Riemann sums, we essentially approximate the area by breaking it into rectangles. The height of each rectangle is determined by the function value at a chosen point within each interval. In this problem, we aim to find an exact area by dividing the interval into an increasing number of rectangles, making them infinitely thin, which leads to an exact calculation of the area.
- We start by setting up a Riemann sum, which is a sum of areas of these rectangles.
- As we increase the number of rectangles (\(n\)), the width of each rectangle (\(\Delta x\)) approaches zero.
- When \(n\to\infty\), the sum converges to a definite integral, representing the exact area under the curve from \(a\) to \(b\).
Right-Hand Endpoint
The right-hand endpoint method is a specific technique used in calculating Riemann sums. In this method, the function value at the right end of each subinterval is used to determine the height of the rectangle. This choice of endpoint can affect the accuracy of the approximation, especially for functions that change rapidly.
- The right-hand endpoint for the k-th subinterval is given by \(x_k = a + k\Delta x\).
- This results in each rectangle capturing the function value at the boundary, ensuring the interval precisely reaches the right endpoint.
- In our example, for the function \(f(x) = x^2 + 1\) on \([0,3]\), right-hand endpoints are used to set up the sum, improving the approximation as \(n\) increases.
Interval Division
Interval division refers to the process of breaking down the overall interval into smaller, equal sections. This is a necessary step in setting up a Riemann sum, as these divisions determine how many rectangles will be used in the approximation.
- The length of each subinterval, \(\Delta x\), is calculated by \(\Delta x = \frac{b-a}{n}\).
- In the case of our interval \([0, 3]\), dividing it into \(n\) intervals gives \(\Delta x = \frac{3}{n}\).
- Dividing into more intervals results in thinner rectangles, providing a better approximation.
Area Under the Curve
The main goal of using Riemann sums in calculus is to find the area under a curve between two points, denoted here by \([a, b]\). This concept is fundamental to integral calculus and has wide applications in physics, engineering, and beyond.
- By utilizing Riemann sums, we approximate this area through the summation of rectangles under the curve.
- As \(n\) approaches infinity, this approximation converges to the exact area, expressed as \(\int_a^b f(x)\,dx\).
- For the specific example of \(f(x) = x^2 + 1\) over \([0,3]\), taking the Riemann sum limit as \(n \to \infty\) yields \(12\), representing the exact area under the curve.
