Explanation Given: F ' ( x ) = f ( x ) F ( 0 ) = 3 F ( 2 ) = 7 Formula Used: A r e a = ∫ a b f ( x ) d x , [ a , b ] = F ( b ) − F ( a ) Calculation: (a) The area under y = f ( x ) over [ 0 , 2 ] , f ( x ) ≥ 0 is ∫ 0 2 f ( x ) d x ∫ 0 2

4th Edition

ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Riemann Sum

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Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.

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Question

Chapter 5.4, Problem 1PQ

To determine

(a) The area under $y = f (x)$ over $[0, 2]$ if $f (x) \geq 0$

(b) The graphical representation of $F (2) - F (0)$ if $f (x)$ takes on both positive and negative values.

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See solution

Chapter 5.3, Problem 86E

Chapter 5.4, Problem 2PQ

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Chapter 5 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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(a) The area under y = f ( x ) over [ 0 , 2 ] if f ( x ) ≥ 0 (b) The graphical representation of F ( 2 ) − F ( 0 ) if f ( x ) takes on both positive and negative values.

Solution Summary: The author calculates the area under y = f ( x ) over [ 0, 2 ] if he takes both positive and negative values.

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(a) The area under $y = f (x)$ over $[0, 2]$ if $f (x) \geq 0$

(b) The graphical representation of $F (2) - F (0)$ if $f (x)$ takes on both positive and negative values.

Want to see the full answer?

Chapter 5 Solutions

(a) The area under y = f ( x ) over [ 0 , 2 ] if f ( x ) ≥ 0 (b) The graphical representation of F ( 2 ) − F ( 0 ) if f ( x ) takes on both positive and negative values.

Solution Summary: The author calculates the area under y = f ( x ) over [ 0, 2 ] if he takes both positive and negative values.

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(a) The area under y=f(x)over [0,2]if f(x)≥0 (b) The graphical representation of F(2)−F(0)if f(x)takes on both positive and negative values.

Want to see the full answer?

Chapter 5 Solutions

(a) The area under $y = f (x)$ over $[0, 2]$ if $f (x) \geq 0$

(b) The graphical representation of $F (2) - F (0)$ if $f (x)$ takes on both positive and negative values.