Explanation Given: The property to be verified is, ∫ a b k f ( x ) d x = k ∫ a b f ( x ) d x Where f ( x ) is integrable on [ a , b ] and k is a constant. Formula used: We know that area bounded by the graph of f ( x ) and x -axis on interval [ a , b ] is represented by, ∫ a b f ( x ) d x Proof: Take a function f ( x ) = 3 x which is integrable on [ 1 , 2 ] First we will draw the graph of f ( x ) = 3 x for that substitute x = 1 , 2 in f ( x ) . At x = 1 , f ( 1 ) = 3 ( 1 ) = 3 And, At x = 2 f ( 2 ) = 3 ( 2 ) = 6 Any line can be drawn using two points and the points are ( 1 , 3 ) , ( 2 , 6 ) plot the points on the coordinate axes and shade the region to obtain, This figure is a trapezoid of altitude 1 and the parallel bases of 3 and 6 . So, ∫ 1 2 ( 3 x ) d x = area of trapezoid . Using, ( area of trapezoid ) = 1 2 h ( a + b ) to get, ∫ 1 2 ( 3 x ) d x = 1 2 ( 1 ) ( 3 + 6 ) = 9 2 Now, draw the graph of f ( x ) = x for that substitute x = 1 , 2 in f ( x )

Calculus Of A Single Variable With Calcchat And Calcview, 11e

11th Edition

ISBN: 9781337275583

Author: Larson

Publisher: Cengage Learning

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Chapter 4.3, Problem 56E

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To prove: The definite integrals ∫abkf(x)dx=k∫abf(x) dx.

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Chapter 4.3, Problem 55E

Chapter 4.3, Problem 57E

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Calculus Of A Single Variable With Calcchat And Calcview, 11e

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To prove: The definite integrals ∫ a b k f ( x ) d x = k ∫ a b f ( x ) d x .

Solution Summary: The author explains that f(x)=3x is integrable on left[a,b], and k a constant.

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To prove: The definite integrals ∫abkf(x)dx=k∫abf(x) dx.

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Chapter 4 Solutions