Evaluate the definite integral and interpret the result. (x3 - 4x) dx i the area between the x-axis and the graph of y -x3 - 4x over the interval (-1, 0] minus the area between the x-axis and the graph of y=x3 - 4x over the interval (0, 2] is - 2. 3; the area between the x-axis and the graph of y = x³ - 4x over the interval (-1, 0] minus the area between the x-axis and the graph of y=x3 - 4x over the interval (0, 2] is 3. 9; the area under the graph of y = x³ - 4x over the interval [-1, 2] is 9. the area between the x-axis and the graph of y = x3 - 4x over the interval [-1, 2] is -.
Evaluate the definite integral and interpret the result. (x3 - 4x) dx i the area between the x-axis and the graph of y -x3 - 4x over the interval (-1, 0] minus the area between the x-axis and the graph of y=x3 - 4x over the interval (0, 2] is - 2. 3; the area between the x-axis and the graph of y = x³ - 4x over the interval (-1, 0] minus the area between the x-axis and the graph of y=x3 - 4x over the interval (0, 2] is 3. 9; the area under the graph of y = x³ - 4x over the interval [-1, 2] is 9. the area between the x-axis and the graph of y = x3 - 4x over the interval [-1, 2] is -.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 67E
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Evaluate the definite
![Evaluate the definite integral and interpret the result.
(x3 - 4x) dx
-1
-6 -5 -4 -3 12
3 45 6x
-; the area between the x-axis and the graph of y=x3 - 4x over the interval [-1, 0] minus the area between the x-axis and the graph of
y=x3 - 4x over the interval [0, 2] is -
3; the area between the x-axis and the graph of y =x - 4x over the interval [-1, 0] minus the area between the x-axis and the graph of
=x3 - 4x over the interval [0, 2] is 3.
%3D
9; the area under the graph of y =x3 - 4x over the interval [-1, 2] is 9.
the area between the x-axis and the graph of y =x3 - 4x over the interval [-1, 2] is -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7bafd92b-9abb-4bf4-8a95-c93ef0194bb1%2F2395e5e6-57ea-4f40-8730-7bad9972b669%2Fktjfioa_processed.png&w=3840&q=75)
Transcribed Image Text:Evaluate the definite integral and interpret the result.
(x3 - 4x) dx
-1
-6 -5 -4 -3 12
3 45 6x
-; the area between the x-axis and the graph of y=x3 - 4x over the interval [-1, 0] minus the area between the x-axis and the graph of
y=x3 - 4x over the interval [0, 2] is -
3; the area between the x-axis and the graph of y =x - 4x over the interval [-1, 0] minus the area between the x-axis and the graph of
=x3 - 4x over the interval [0, 2] is 3.
%3D
9; the area under the graph of y =x3 - 4x over the interval [-1, 2] is 9.
the area between the x-axis and the graph of y =x3 - 4x over the interval [-1, 2] is -
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