Problem 37
Question
Without doing any calculations, rank from smallest to largest the approximations of \(\int_{1}^{6} \frac{1}{x} d x\) for the following methods: left Riemann sum, midpoint Riemann sum, Trapezoidal rule.
Step-by-Step Solution
Verified Answer
Left Riemann sum > Trapezoidal rule > Midpoint Riemann sum.
1Step 1: Understand Each Approximation Method
The integral \(\int_{1}^{6} \frac{1}{x} \, dx\) can be approximated using numerical integration methods like the left Riemann sum, midpoint Riemann sum, and the trapezoidal rule. Each method uses different points in the subintervals to approximate the area under the curve, which can affect their accuracy.
2Step 2: Evaluate Bias of the Left Riemann Sum
The left Riemann sum uses the left endpoints of subintervals to approximate the integral. For the function \(\frac{1}{x}\), which is decreasing over the interval \([1, 6]\), this results in an overestimation because left points have higher function values than right.
3Step 3: Consider the Midpoint Riemann Sum
The midpoint Riemann sum uses the midpoints of subintervals for function values. Given that \(\frac{1}{x}\) is decreasing, the midpoint is typically more centered and hence would provide estimates that are usually closer to true integral values compared to the left Riemann sum.
4Step 4: Analyze the Trapezoidal Rule
The trapezoidal rule averages the values at the endpoints of the subintervals, which accounts for both overestimation at the left and underestimation at the right. As a result, it generally provides a better approximation than the purely left-based approximation methods for decreasing functions, like \(\frac{1}{x}\).
Key Concepts
Left Riemann SumMidpoint Riemann SumTrapezoidal Rule
Left Riemann Sum
The Left Riemann Sum is a numerical integration technique that approximates the integral of a function over a closed interval. This method involves using the left endpoint of each subinterval to estimate the height of rectangles beneath a curve.
For a decreasing function like \( \frac{1}{x} \), this approach will typically result in an overestimate, since the left endpoints are higher on the curve compared to corresponding right endpoints.
This approach is simple but can lead to inaccuracies if the function varies significantly across subintervals. When considering larger intervals or functions that exhibit considerable fluctuation, adjustments or additional calculations may be necessary to improve accuracy.
For a decreasing function like \( \frac{1}{x} \), this approach will typically result in an overestimate, since the left endpoints are higher on the curve compared to corresponding right endpoints.
This approach is simple but can lead to inaccuracies if the function varies significantly across subintervals. When considering larger intervals or functions that exhibit considerable fluctuation, adjustments or additional calculations may be necessary to improve accuracy.
Midpoint Riemann Sum
The Midpoint Riemann Sum is another approximate method, but it uses midpoints of subintervals as a basis for estimating the height of rectangles below the curve. This typically offers more precise estimates than some other methods.
- By using the midpoint, this sum balances out over- and underestimations, leading to results that are often closer to the actual integral value.
- For a decreasing function like \( \frac{1}{x} \), the midpoint provides a more central value, smoothing out extremes of left or right endpoints.
Trapezoidal Rule
The Trapezoidal Rule is a more sophisticated numerical method that approximates integrals by leveraging both endpoints of each subinterval. It essentially uses trapezoids instead of rectangles to estimate the area beneath a curve.
This method calculates the integral's approximation by averaging the function's value at the left and right endpoints, making it particularly effective for functions that decrease over the interval like \( \frac{1}{x} \).
This method calculates the integral's approximation by averaging the function's value at the left and right endpoints, making it particularly effective for functions that decrease over the interval like \( \frac{1}{x} \).
- The trapezoidal rule tends to yield more accurate results compared to the left Riemann sum when dealing with decreasing functions.
- It also balances errors that arise from over- and underestimation, often resulting in better alignment with the actual integral value.
Other exercises in this chapter
Problem 36
Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-1}^{0} \sqrt{x^{3}+1}\left(3 x^{2}\right) d x $$
View solution Problem 37
Many computer algebra systems permit the evaluation of Riemann sums for left end point, right end point, or midpoint evaluations of the function. Using such a s
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Let \(x_{1}, x_{2}, \ldots, x_{n}\) be any real numbers. Find the value of \(c\) that minimizes \(\sum_{i=1}^{n}\left(x_{i}-c\right)^{2}\).
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Use the Substitution Rule for Definite Integrals to evaluate each definite integral. $$ \int_{-1}^{3} \frac{1}{(t+2)^{2}} d t $$
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