[ (2x² - 6x +) dx and interpret the result EXAMPLE 4 Find in terms of areas. SOLUTION The Fundamental Theorem gives [ (2x* - 6x + ) ak = 2() - ) • tan" 3 x2 + 1 dx = + 3 tan- - 3x2 + 3 tan-1 =글(2)-3(27) + 3 tan-I(2) - | + 3 tan-(2). This is the exact value of the integral. If a decimal approximation is desired, we can use a calculator to approximate tan-(2). Doing so, we get 3 - 6x + |dx = . (Round your answer x2 + 1 to four decimal places.) The figure below shows the graph of the integrand. We know that the value of the integral can be interpreted as a net area: the sum of the areas labeled with a plus sign minus the area labeled with a minus sign. +,

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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EXAMPLE 4
Find
6х +
dx and interpret the result
x² + 1
in terms of areas.
SOLUTION
The Fundamental Theorem gives
2() - G)
- 6x +
dx =
+ 3 tan-1
x2 + 1
3x2 + 3 tan
극(24)-3(22) + 3 tan-1(2) -
+ 3 tan-1(2).
This is the exact value of the integral. If a decimal approximation is
desired, we can use a calculator to approximate tan-(2). Doing so, we
get
- 6x +) ax = (
3
(Round your answer
x² + 1
to four decimal places.)
The figure below shows the graph of the integrand. We know that the
value of the integral can be interpreted as a net area: the sum of the
areas labeled with a plus sign minus the area labeled with a minus sign.
y
3
X
2
Transcribed Image Text:EXAMPLE 4 Find 6х + dx and interpret the result x² + 1 in terms of areas. SOLUTION The Fundamental Theorem gives 2() - G) - 6x + dx = + 3 tan-1 x2 + 1 3x2 + 3 tan 극(24)-3(22) + 3 tan-1(2) - + 3 tan-1(2). This is the exact value of the integral. If a decimal approximation is desired, we can use a calculator to approximate tan-(2). Doing so, we get - 6x +) ax = ( 3 (Round your answer x² + 1 to four decimal places.) The figure below shows the graph of the integrand. We know that the value of the integral can be interpreted as a net area: the sum of the areas labeled with a plus sign minus the area labeled with a minus sign. y 3 X 2
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