Example 6 Video Example Use the Midpoint Rule with n = 5 to approximate the following integral. [²3/0x - dx Solution 6 7 8 9 The endpoints of the subintervals are 1, (See the figure below.) 5'5'5'5 and 2, so the midpoints are 1.1, 1.3, 1.5, 1.7, and Δ x = 0.2 C ● 1.2 1.4 1.8 A x 1.1 1.3 1.5 The width of the subintervals is Ax= - (2-¹) - [ so the Midpoint Rule gives the following. ²3 dx = Ax [f(1.1) + f(1.3) + f(1.5) + f(1.7) + f(1.9)] G + 13+15+17+19) (Round your answer to four decimal places.) Since f(x) = > 0 for 1 ≤ x ≤ 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in the figure below. y 3 y = 1.6 ● 1.7 1.9
Example 6 Video Example Use the Midpoint Rule with n = 5 to approximate the following integral. [²3/0x - dx Solution 6 7 8 9 The endpoints of the subintervals are 1, (See the figure below.) 5'5'5'5 and 2, so the midpoints are 1.1, 1.3, 1.5, 1.7, and Δ x = 0.2 C ● 1.2 1.4 1.8 A x 1.1 1.3 1.5 The width of the subintervals is Ax= - (2-¹) - [ so the Midpoint Rule gives the following. ²3 dx = Ax [f(1.1) + f(1.3) + f(1.5) + f(1.7) + f(1.9)] G + 13+15+17+19) (Round your answer to four decimal places.) Since f(x) = > 0 for 1 ≤ x ≤ 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in the figure below. y 3 y = 1.6 ● 1.7 1.9
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.3: The Addition And Subtraction Formulas
Problem 76E
Related questions
Question
![Example 6
Video Example)
Use the Midpoint Rule with n = 5 to approximate the following integral.
[²3/1 0
Solution
6 7 8 9
. (See the figure below.)
The endpoints of the subintervals are 1,555 and 2, so the midpoints are 1.1, 1.3, 1.5, 1.7, and
Ax=0.2
1.2
1.4
1.6 A 1.8
x
1.1
1.3
1.5
1.7
(2-1)
The width of the subintervals is Ax = =
= [
so the Midpoint Rule gives the following.
[²3/1dx
dx ≈ Ax [f(1.1) + f(1.3) + f(1.5) + f(1.7) +
) + f(1.9)]
f
= (₁/1
+
237 +233)
1.3 1.5
1.7
(Round your answer to four decimal places.)
Since f(x) = -> 0 for 1 ≤ x ≤ 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in the figure below.
y
y =
dx
+
1.9](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce60c797-cede-4a9b-a194-cc1528c2f13e%2F189908cd-35e3-42fc-960e-3b6f7518a70a%2Fz4iy5l_processed.png&w=3840&q=75)
Transcribed Image Text:Example 6
Video Example)
Use the Midpoint Rule with n = 5 to approximate the following integral.
[²3/1 0
Solution
6 7 8 9
. (See the figure below.)
The endpoints of the subintervals are 1,555 and 2, so the midpoints are 1.1, 1.3, 1.5, 1.7, and
Ax=0.2
1.2
1.4
1.6 A 1.8
x
1.1
1.3
1.5
1.7
(2-1)
The width of the subintervals is Ax = =
= [
so the Midpoint Rule gives the following.
[²3/1dx
dx ≈ Ax [f(1.1) + f(1.3) + f(1.5) + f(1.7) +
) + f(1.9)]
f
= (₁/1
+
237 +233)
1.3 1.5
1.7
(Round your answer to four decimal places.)
Since f(x) = -> 0 for 1 ≤ x ≤ 2, the integral represents an area, and the approximation given by the Midpoint Rule is the sum of the areas of the rectangles shown in the figure below.
y
y =
dx
+
1.9
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