Use the Midpoint Rule with n = 5 to approximate the following integral. [² x Solution The endpoints of the subintervals are 1, 1 1.2 1.4 RE 1.6 3 Ax=0.4 1.8 The width of the subintervals is Ax = 7 9 11 13 2 2.2 3 3 = - 3 ( 1²/2² - 16-2²-24-2²6) = (3-¹) = [ dx ≈ Ax[R(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)] 2.4 and 3, so the midpoints are 1.2, 1.5, 2, 2.4, and [ 2.64 3 2.8 3 x ], so the Midpoint Rule gives the following. .(Round your answer to four decimal places.) Since f(x) = -²x > 0 for 1 ≤ x ≤ 3, 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. . (See the figure below.)
Use the Midpoint Rule with n = 5 to approximate the following integral. [² x Solution The endpoints of the subintervals are 1, 1 1.2 1.4 RE 1.6 3 Ax=0.4 1.8 The width of the subintervals is Ax = 7 9 11 13 2 2.2 3 3 = - 3 ( 1²/2² - 16-2²-24-2²6) = (3-¹) = [ dx ≈ Ax[R(1.2) + f(1.6) + f(2) + f(2.4) + f(2.8)] 2.4 and 3, so the midpoints are 1.2, 1.5, 2, 2.4, and [ 2.64 3 2.8 3 x ], so the Midpoint Rule gives the following. .(Round your answer to four decimal places.) Since f(x) = -²x > 0 for 1 ≤ x ≤ 3, 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. . (See the figure below.)
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.2: Integration By Parts
Problem 34E
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![Use the Midpoint Rule with n= 5 to approximate the following integral.
[²3/1 ax
dx
Solution
The endpoints of the subintervals are 1,
1
1.2
1.4
y
1.6
Ax=0.4
1.8
1
7 9 11 13
5' 5' 5' 5
2
2/
3
3
=
- 3(2³2+36 +2-24-28)
1.6
The width of the subintervals is 4x = (3 − ¹) = [
3 dx = 4x [R(1.2) + f(1.6) + F(2) + F(2.4) + f(2.8)]
3
X
2.2 A
3
2.4
and 3, so the midpoints are 1.2, 1.6, 2, 2.4, and
2.6 A
3
2.8
+
3
. (Round your answer to four decimal places.)
Since f(x) = 3 >
> 0 for 1 ≤ x ≤ 3, 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.
x
Q
], so the Midpoint Rule gives the following.
. (See the figure below.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe2f84c35-2f72-4af0-b5c7-1dcb9cfb2313%2Fd015296b-2a1c-47ca-bcc9-e613d80f4f2d%2F56s14o_processed.png&w=3840&q=75)
Transcribed Image Text:Use the Midpoint Rule with n= 5 to approximate the following integral.
[²3/1 ax
dx
Solution
The endpoints of the subintervals are 1,
1
1.2
1.4
y
1.6
Ax=0.4
1.8
1
7 9 11 13
5' 5' 5' 5
2
2/
3
3
=
- 3(2³2+36 +2-24-28)
1.6
The width of the subintervals is 4x = (3 − ¹) = [
3 dx = 4x [R(1.2) + f(1.6) + F(2) + F(2.4) + f(2.8)]
3
X
2.2 A
3
2.4
and 3, so the midpoints are 1.2, 1.6, 2, 2.4, and
2.6 A
3
2.8
+
3
. (Round your answer to four decimal places.)
Since f(x) = 3 >
> 0 for 1 ≤ x ≤ 3, 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.
x
Q
], so the Midpoint Rule gives the following.
. (See the figure below.)
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