Problem 136
Question
[T] Compute the left and right Riemann sums \(L_{10}\) and \(R_{10}\) and their average \(\frac{L_{10}+R_{10}}{2}\) for \(f(t)=t^{2}\) over \([0,1] .\) Given that \(\int_{0}^{1} t^{2} d t=0 . \overline{33}, \quad\) to how many decimal places is \(\frac{L_{10}+R_{10}}{2}\) accurate?
Step-by-Step Solution
Verified Answer
The average, 0.335, is accurate to 1 decimal place.
1Step 1: Determine the interval width
The interval \([0,1]\) is divided into 10 sub-intervals for both the left and right Riemann sums. Thus, the width of each sub-interval \( \Delta t \) is given by: \[ \Delta t = \frac{1 - 0}{10} = 0.1 \]
2Step 2: Compute Left Riemann Sum \(L_{10}\)
For the left Riemann sum \(L_{10}\), we evaluate \(f(t) = t^2\) at the left endpoints of each sub-interval. The left endpoints are \(t_0 = 0, t_1 = 0.1, ..., t_9 = 0.9\). The sum is: \[ L_{10} = 0.1 \left[ f(0) + f(0.1) + \cdots + f(0.9) \right] \]Calculating each term: \[ L_{10} = 0.1 \left[ 0^2 + 0.1^2 + 0.2^2 + \cdots + 0.9^2 \right] = 0.1 (0 + 0.01 + 0.04 + 0.09 + 0.16 + 0.25 + 0.36 + 0.49 + 0.64 + 0.81) \]\[= 0.1 \times 2.85 = 0.285 \]
3Step 3: Compute Right Riemann Sum \(R_{10}\)
For the right Riemann sum \(R_{10}\), we evaluate \(f(t) = t^2\) at the right endpoints of each sub-interval. The right endpoints are \(t_1 = 0.1, t_2 = 0.2, ..., t_{10} = 1.0\). The sum is: \[ R_{10} = 0.1 \left[ f(0.1) + f(0.2) + \cdots + f(1.0) \right] \]Calculating each term: \[ R_{10} = 0.1 \left[ 0.1^2 + 0.2^2 + 0.3^2 + \cdots + 1.0^2 \right] = 0.1 (0.01 + 0.04 + 0.09 + 0.16 + 0.25 + 0.36 + 0.49 + 0.64 + 0.81 + 1.00) \]\[= 0.1 \times 3.85 = 0.385 \]
4Step 4: Compute the Average of \(L_{10}\) and \(R_{10}\)
The average of the left and right Riemann sums is: \[ \frac{L_{10} + R_{10}}{2} = \frac{0.285 + 0.385}{2} = \frac{0.67}{2} = 0.335 \]
5Step 5: Compare with the Exact Integral Value
The exact value of the integral is given by \( \int_{0}^{1} t^2 dt = 0.\overline{3} \). Comparing with our average, \( 0.335 \), we see they agree to: - The first decimal place, since their difference is less than 0.01.
Key Concepts
Left Riemann SumRight Riemann SumAverage of Riemann Sums
Left Riemann Sum
The left Riemann sum provides an approximation of the area under the curve of a function over a specified interval. It relies on evaluating the function at the left endpoints of a set number of equal sub-intervals within the main interval.
For the exercise given, the function we are working with is \(f(t) = t^2\) over the interval \([0, 1]\). Our task was to divide it into 10 equal parts, where each sub-interval width, denoted \(\Delta t\), is calculated as \(0.1\).
In computing the left Riemann sum \(L_{10}\), start by evaluating the function at these left endpoints: \( t_0 = 0, t_1 = 0.1, ... , t_9 = 0.9 \). These values form the basis for constructing the sum, which is \[ L_{10} = 0.1 \left[ f(0) + f(0.1) + \cdots + f(0.9) \right] \].
By substituting each \(f(t)\) value into \( (t^2)\), then summing the results, we get:\[ L_{10} = 0.1 \times 2.85 = 0.285 \]. Hence, the left Riemann sum gives us this approximation.
For the exercise given, the function we are working with is \(f(t) = t^2\) over the interval \([0, 1]\). Our task was to divide it into 10 equal parts, where each sub-interval width, denoted \(\Delta t\), is calculated as \(0.1\).
In computing the left Riemann sum \(L_{10}\), start by evaluating the function at these left endpoints: \( t_0 = 0, t_1 = 0.1, ... , t_9 = 0.9 \). These values form the basis for constructing the sum, which is \[ L_{10} = 0.1 \left[ f(0) + f(0.1) + \cdots + f(0.9) \right] \].
By substituting each \(f(t)\) value into \( (t^2)\), then summing the results, we get:\[ L_{10} = 0.1 \times 2.85 = 0.285 \]. Hence, the left Riemann sum gives us this approximation.
Right Riemann Sum
The right Riemann sum, in contrast to the left Riemann sum, evaluates the function using the right endpoints of sub-intervals. This provides another technique to estimate the integral of the function across a designated interval.
In the exercise, we deal with the function \( f(t) = t^2 \) and the interval \([0, 1]\), divided into 10 slices, each of width \(0.1\). For \(R_{10}\), compute the function at each right endpoint: \( t_1 = 0.1, t_2 = 0.2, ..., t_{10} = 1.0 \). With this setup, our sum is:\[ R_{10} = 0.1 \left[ f(0.1) + f(0.2) + \cdots + f(1.0) \right] \].
Substituting \( f(t) = t^2 \) and evaluating each term, the result is: \[ R_{10} = 0.1 \times 3.85 = 0.385 \]. Therefore, the right Riemann sum calculates the area under curve as 0.385, providing slightly different insights compared to the left sum.
In the exercise, we deal with the function \( f(t) = t^2 \) and the interval \([0, 1]\), divided into 10 slices, each of width \(0.1\). For \(R_{10}\), compute the function at each right endpoint: \( t_1 = 0.1, t_2 = 0.2, ..., t_{10} = 1.0 \). With this setup, our sum is:\[ R_{10} = 0.1 \left[ f(0.1) + f(0.2) + \cdots + f(1.0) \right] \].
Substituting \( f(t) = t^2 \) and evaluating each term, the result is: \[ R_{10} = 0.1 \times 3.85 = 0.385 \]. Therefore, the right Riemann sum calculates the area under curve as 0.385, providing slightly different insights compared to the left sum.
Average of Riemann Sums
The average of the left and right Riemann sums offers a refined approximation of the area under the curve. By averaging \(L_{10}\) and \(R_{10}\), we balance any discrepancies that might arise from solely using the left or right estimates.
In the context of this problem, we calculated \(L_{10}\) and \(R_{10}\) as 0.285 and 0.385, respectively. To find their average, use: \[ \frac{L_{10} + R_{10}}{2} = \frac{0.285 + 0.385}{2} = 0.335 \]
This averaged result, 0.335, is checked against the exact integral \( \int_{0}^{1} t^2 dt = 0.\overline{33} \). The comparison demonstrates a close match, being accurate to the first decimal place, which encourages confidence in employing these approximation tools when finding integrals.
In the context of this problem, we calculated \(L_{10}\) and \(R_{10}\) as 0.285 and 0.385, respectively. To find their average, use: \[ \frac{L_{10} + R_{10}}{2} = \frac{0.285 + 0.385}{2} = 0.335 \]
This averaged result, 0.335, is checked against the exact integral \( \int_{0}^{1} t^2 dt = 0.\overline{33} \). The comparison demonstrates a close match, being accurate to the first decimal place, which encourages confidence in employing these approximation tools when finding integrals.
Other exercises in this chapter
Problem 134
Suppose that for each \(i\) such that \(1 \leq i \leq N\) one has \(\int_{i-1}^{i} f(t) d t=i .\) Show that \(\int_{0}^{N} f(t) d t=\frac{N(N+1)}{2}\)
View solution Problem 135
Suppose that for each \(i\) such that \(1 \leq i \leq N\) one has \(\int_{i-1}^{i} f(t) d t=i^{2}\) show that \(\int_{0}^{N} f(t) d t=\frac{N(N+1)(2 N+1)}{6}\).
View solution Problem 137
[T] Compute the left and right Riemann sums, \(L_{10}\) and \(R_{10},\) and their average \(\frac{L_{10}+R_{10}}{2}\) for \(f(t)=\left(4-t^{2}\right)\) over \([
View solution Problem 138
If \( \int_{1}^{5} \sqrt{1+t^{4}} d t=41.7133 \ldots, \) what is \(\int_{1}^{5} \sqrt{1+u^{4}} d u ?\)
View solution