Problem 45
Question
Riemann sums for larger values of \(n\) Complete the following steps for the given function \(f\) and interval. a. For the given value of \(n\), use sigma notation to write the left, right, and midpoint Riemann sums. Then evaluate each sum using a calculator. b. Based on the approximations found in part (a), estimate the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the interval. $$f(x)=x^{2}-1 \text { on }[2,7] ; n=75$$
Step-by-Step Solution
Verified Answer
Question:
Estimate the area of the region bounded by the graph of the function \(f(x) = x^2 - 1\) and the \(x\)-axis on the interval \([2,7]\) using left, right, and midpoint Riemann sums with \(n = 75\) subintervals.
Answer:
To estimate the area, follow these steps:
1. Calculate the width of each subinterval, \(\Delta x = \frac{1}{15}\).
2. Calculate the left Riemann sum, \(L_n = \frac{1}{15}\sum_{i=0}^{74} \left[ (2+\frac{i}{15})^2 - 1\right]\).
3. Calculate the right Riemann sum, \(R_n = \frac{1}{15}\sum_{i=1}^{75} \left[ (2+\frac{i}{15})^2 - 1\right]\).
4. Calculate the midpoint Riemann sum, \(M_n = \frac{1}{15}\sum_{i=0}^{74} \left[ \left(2+\frac{1}{30} +\frac{i}{15}\right)^2 - 1 \right]\).
5. Estimate the area using the average of the left, right, and midpoint Riemann sums: \(Area \approx \frac{L_n+R_n+M_n}{3}\).
Use a calculator to find the values of \(L_n\), \(R_n\), and \(M_n\) and substitute them into the area estimation formula.
1Step 1: Calculate the width of subintervals
We first need to divide the interval \([2, 7]\) into \(n=75\) equal subintervals. To calculate the width of each subinterval \(\Delta x\), we use the formula:
\(\Delta x = \frac{b-a}{n}\)
where \(a\) is the starting point of the interval, \(b\) is the ending point of the interval, and \(n\) is the number of subintervals.
In our case, \(a=2\), \(b=7\), and \(n=75\). So the width of each subinterval is:
\(\Delta x = \frac{7-2}{75} = \frac{5}{75}= \frac{1}{15}\).
2Step 2: Left Riemann Sum
For the left Riemann sum, we need to evaluate the function at the left endpoint of each subinterval and then sum these results, multiplied by the width of the subintervals:
$$L_n = \Delta x\sum_{i=0}^{n-1} f(a+i\Delta x)$$
Substitute the given function and the calculated \(\Delta x\) into the formula:
$$L_n = \frac{1}{15}\sum_{i=0}^{74} \left[ (2+\frac{i}{15})^2 - 1\right]$$
Now, evaluate this sum using a calculator.
3Step 3: Right Riemann Sum
For the right Riemann sum, we need to evaluate the function at the right endpoint of each subinterval and then sum these results, multiplied by the width of the subintervals:
$$R_n = \Delta x\sum_{i=1}^{n} f(a+i\Delta x)$$
Substitute the given function and the calculated \(\Delta x\) into the formula:
$$R_n = \frac{1}{15}\sum_{i=1}^{75} \left[ (2+\frac{i}{15})^2 - 1\right]$$
Now, evaluate this sum using a calculator.
4Step 4: Midpoint Riemann Sum
For the midpoint Riemann sum, we need to evaluate the function at the midpoint of each subinterval and then sum up these results, multiplied by the width of the subintervals:
$$M_n = \Delta x\sum_{i=0}^{n-1} f\left(a+\frac{1}{2}\Delta x+(i\Delta x)\right)$$
Substitute the given function and the calculated \(\Delta x\) into the formula:
$$M_n = \frac{1}{15}\sum_{i=0}^{74} \left[ \left(2+\frac{1}{30} +\frac{i}{15}\right)^2 - 1 \right]$$
Now, evaluate this sum using a calculator.
5Step 5: Estimate the area
Based on the results from steps 2, 3, and 4, we will now estimate the area of the region bounded by the graph of \(f\) and the \(x\)-axis on the interval \([2,7]\):
$$Area \approx \frac{L_n+R_n+M_n}{3}$$
Substitute the values of \(L_n\), \(R_n\), and \(M_n\) obtained using a calculator in the above formula to find the estimated area.
Key Concepts
Sigma NotationSubintervalLeft Riemann SumRight Riemann SumMidpoint Riemann SumApproximation of AreaFunction Evaluation
Sigma Notation
Sigma notation is a compact way to represent the sum of a series of terms. When dealing with Riemann sums, it's especially useful because it helps you express the sum of function evaluations over multiple subintervals with ease. In its general form, sigma notation looks like this:
\[\sum_{i=m}^{n} a_i\]
This means that for each value from \(m\) to \(n\), you should add up the values of \(a_i\). When seen in the context of Riemann sums, sigma notation helps handle numerous function evaluations without writing them all out individually. This makes it much easier to work with any partition size, especially if the number of subintervals \(n\) is large, as in our example where \(n = 75\).
\[\sum_{i=m}^{n} a_i\]
This means that for each value from \(m\) to \(n\), you should add up the values of \(a_i\). When seen in the context of Riemann sums, sigma notation helps handle numerous function evaluations without writing them all out individually. This makes it much easier to work with any partition size, especially if the number of subintervals \(n\) is large, as in our example where \(n = 75\).
Subinterval
To approximate the area under a curve using Riemann sums, the interval \[a, b\]\ is divided into smaller sections, known as subintervals. The length of each subinterval is usually denoted by \(\Delta x\), and it's calculated based on the total length of \[a, b\]\ and the number of subintervals \(n\). The formula for finding the width of each subinterval is:
\[\Delta x = \frac{b-a}{n}\]
In our example, since the interval \[2, 7\]\ is divided into 75 subintervals, each subinterval has a width of \(\frac{1}{15}\). Splitting the interval into these smaller sections allows us to systematically evaluate the function across the entire range from \a\ to \b\.
\[\Delta x = \frac{b-a}{n}\]
In our example, since the interval \[2, 7\]\ is divided into 75 subintervals, each subinterval has a width of \(\frac{1}{15}\). Splitting the interval into these smaller sections allows us to systematically evaluate the function across the entire range from \a\ to \b\.
Left Riemann Sum
The left Riemann sum is one of three common types of Riemann sums used for approximating the area under a curve. With the left Riemann sum, you evaluate the function at the left endpoint of each subinterval. This means that for each subinterval, you use the starting point to estimate the area of that slice.
Here's the formula for the left Riemann sum:\[L_n = \Delta x \sum_{i=0}^{n-1} f(a + i\Delta x)\]
This method can lead to an underestimation or overestimation depending on whether the function is increasing or decreasing within the interval. For the given function \(f(x) = x^2 - 1\), calculating the left Riemann sum involves summing up evaluations at each left endpoint from 2 through just before 7.
Here's the formula for the left Riemann sum:\[L_n = \Delta x \sum_{i=0}^{n-1} f(a + i\Delta x)\]
This method can lead to an underestimation or overestimation depending on whether the function is increasing or decreasing within the interval. For the given function \(f(x) = x^2 - 1\), calculating the left Riemann sum involves summing up evaluations at each left endpoint from 2 through just before 7.
Right Riemann Sum
The right Riemann sum uses the opposite approach to the left Riemann sum by evaluating the function at the right endpoint of each subinterval. Its formula is:
\[R_n = \Delta x \sum_{i=1}^{n} f(a + i\Delta x)\]
This approach sometimes yields different approximations than the left Riemann sum. If the function is increasing, then the right sum might overestimate the area. Conversely, if the function is decreasing, it might underestimate it. When computing the right Riemann sum for our function \(f(x) = x^2 - 1\), you evaluate the function starting just to the right of 2, moving in equal increments up to 7.
\[R_n = \Delta x \sum_{i=1}^{n} f(a + i\Delta x)\]
This approach sometimes yields different approximations than the left Riemann sum. If the function is increasing, then the right sum might overestimate the area. Conversely, if the function is decreasing, it might underestimate it. When computing the right Riemann sum for our function \(f(x) = x^2 - 1\), you evaluate the function starting just to the right of 2, moving in equal increments up to 7.
Midpoint Riemann Sum
The midpoint Riemann sum is distinct from the left and right sums. Instead of using the ends of subintervals, it evaluates the function at the midpoint of each subinterval. This is often a better approximation for the area because it balances the over and underestimations of the other two methods.
Here's how you represent the midpoint Riemann sum:\[M_n = \Delta x \sum_{i=0}^{n-1} f\left(a + \frac{1}{2}\Delta x + i\Delta x\right)\]
For our example, you take the midpoint of each subinterval for evaluating \(f(x) = x^2 - 1\), providing a well-rounded estimate by leveraging the central point of each division. This method aims to provide a better approximation, especially when the function behaves non-uniformly within the interval.
Here's how you represent the midpoint Riemann sum:\[M_n = \Delta x \sum_{i=0}^{n-1} f\left(a + \frac{1}{2}\Delta x + i\Delta x\right)\]
For our example, you take the midpoint of each subinterval for evaluating \(f(x) = x^2 - 1\), providing a well-rounded estimate by leveraging the central point of each division. This method aims to provide a better approximation, especially when the function behaves non-uniformly within the interval.
Approximation of Area
Approximating the area under a curve with Riemann sums involves combining the results from different summation methods to form a comprehensive estimate. By comparing the left, right, and midpoint sums, you gain a better understanding of the function’s behavior across the interval.
In practice, you might take the average of these values to estimate the true area, as it considers multiple perspectives of how the function behaves between the endpoints.
In our case, the approximate area under \(f(x) = x^2 - 1\) from 2 to 7 is calculated by averaging the three Riemann sums:
\[\text{Area} \approx \frac{L_n + R_n + M_n}{3}\]
This gives a balanced estimation by smoothing out individual method errors.
In practice, you might take the average of these values to estimate the true area, as it considers multiple perspectives of how the function behaves between the endpoints.
In our case, the approximate area under \(f(x) = x^2 - 1\) from 2 to 7 is calculated by averaging the three Riemann sums:
\[\text{Area} \approx \frac{L_n + R_n + M_n}{3}\]
This gives a balanced estimation by smoothing out individual method errors.
Function Evaluation
Function evaluation is a key step in all types of Riemann sums. It involves substituting the chosen point – whether it’s the left, right, or midpoint of a subinterval – into the function to find its value at that specific point.
For example, if you're using the function \(f(x) = x^2 - 1\), and you pick a point \(x_i\) to evaluate this function at, it involves calculating \((x_i)^2 - 1\).
Repeatedly carrying out this evaluation for each subinterval according to their method (left, right, or midpoint) is necessary to sum up the correct inputs and outputs. With many subintervals like in our case where \(n = 75\), accurate function evaluation is crucial for a precise Riemann sum calculation.
For example, if you're using the function \(f(x) = x^2 - 1\), and you pick a point \(x_i\) to evaluate this function at, it involves calculating \((x_i)^2 - 1\).
Repeatedly carrying out this evaluation for each subinterval according to their method (left, right, or midpoint) is necessary to sum up the correct inputs and outputs. With many subintervals like in our case where \(n = 75\), accurate function evaluation is crucial for a precise Riemann sum calculation.
Other exercises in this chapter
Problem 45
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