Find the area of the region bounded by the curves using the relation: 1= / (f(x) – 9 (x)) dz (2) A = Here, the lower limit is a, the upper limit is b, the top curve function is f (æ), and the bottom curve function is g (æ). Substitute –1 for a, 1 for b, e* for f (x), and a? – 1 for g (x) in Equation (2). = [e² - (2² – 1)] dz A : [e - a? + 1] dz = 2[e - + 2], =2((* - + 1) – (")) =2 (2.7183 -를 + 1-1) %3D =2(11549 2 4.7699 Therefore, the area of the region enclosed by the curves is 4.7699.

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.
Chapter7: Integration
Section7.CR: Chapter 7 Review
Problem 54CR
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where does the 2 that gets pulled out infront of the integral come from?

Find the area of the region bounded by the curves using the relation:
1= / (f(x) – 9 (x)) dz (2)
A =
Here, the lower limit is a, the upper limit is b, the top curve function is
f (æ), and the bottom curve function is g (æ).
Substitute –1 for a, 1 for b, e* for f (x), and a? – 1 for g (x) in Equation
(2).
= [e² - (2² – 1)] dz
A :
[e - a? + 1] dz
= 2[e - + 2],
=2((* - + 1) – ("))
=2 (2.7183 -를 + 1-1)
%3D
=2(11549
2 4.7699
Therefore, the area of the region enclosed by the curves is 4.7699.
Transcribed Image Text:Find the area of the region bounded by the curves using the relation: 1= / (f(x) – 9 (x)) dz (2) A = Here, the lower limit is a, the upper limit is b, the top curve function is f (æ), and the bottom curve function is g (æ). Substitute –1 for a, 1 for b, e* for f (x), and a? – 1 for g (x) in Equation (2). = [e² - (2² – 1)] dz A : [e - a? + 1] dz = 2[e - + 2], =2((* - + 1) – (")) =2 (2.7183 -를 + 1-1) %3D =2(11549 2 4.7699 Therefore, the area of the region enclosed by the curves is 4.7699.
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