Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![5. Using 4 equal-width intervals, show that the trapezoidal rule is the average of the upper and lower sum estimates for 1x²dx.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9537fa2-79e3-471e-bb9d-b59824abcc1f%2F122464a7-f51e-4b62-af65-6ca354de9d6a%2Fx8haj_processed.png&w=3840&q=75)
Transcribed Image Text:5. Using 4 equal-width intervals, show that the trapezoidal rule is the average of the upper and lower sum estimates for 1x²dx.
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Step 1: Given
VIEWStep 2: Formula for trapezoidal rule, upper sum and lower sum
VIEWStep 3: Determine the sub-intervals and function value at each value of x
VIEWStep 4: Determine the integral using trapezoidal rule
VIEWStep 5: Determine the integral using upper sum
VIEWStep 6: Determine the integral using lower sum
VIEWStep 7: Prove that the trapezoidal rule is the average of upper sum and lower sum estimation
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I have a question about the problem solved athttps://www.bartleby.com/questions-and-answers/5.-using-4-equal-width-intervals-show-that-the-trapezoidal-rule-is-the-average-of-the-upper-and-lowe/122464a7-f51e-4b62-af65-6ca354de9d6a
In step 2 shouldn't 2fx3=3, not 4.5?
![Step 2
2
0
L
du
Trapezoidal vale
1h² f(x) de =
Ax =
Xo = 01
។
X 1 = xod AR = 1/₂
F(no)
f(o). O
2F(x₁) = 2f(1/₂) = 1/2 -0.5
2 F (1₂) =
2 F(^₂) =
f(14) =
A² [FR] + 2 f(x₁ + 2 f(12) -12 f(23) -1 ---+
b-a
n
i
2F(1) = 2
2F ( ¾/₂) = 4.5
f(2)=
4
2.75
2
√²x²de = 1/4 [0+0-5+2+4.5+4]
1
12 = 1 1 × 3 = ¾/1₂, xy = 2
- + f(an)]](https://content.bartleby.com/qna-images/question/e9537fa2-79e3-471e-bb9d-b59824abcc1f/757b74ca-defb-4d10-b9e4-da430b0272ff/ihqgcy_thumbnail.png)
Transcribed Image Text:Step 2
2
0
L
du
Trapezoidal vale
1h² f(x) de =
Ax =
Xo = 01
។
X 1 = xod AR = 1/₂
F(no)
f(o). O
2F(x₁) = 2f(1/₂) = 1/2 -0.5
2 F (1₂) =
2 F(^₂) =
f(14) =
A² [FR] + 2 f(x₁ + 2 f(12) -12 f(23) -1 ---+
b-a
n
i
2F(1) = 2
2F ( ¾/₂) = 4.5
f(2)=
4
2.75
2
√²x²de = 1/4 [0+0-5+2+4.5+4]
1
12 = 1 1 × 3 = ¾/1₂, xy = 2
- + f(an)]
Solution
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