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need help solving this calc question

+1 vote

 

Find the average value of f(x) in the interval [-1,3] when f'(x)=3x^2-6x and f(2)=0?

asked Jan 18, 2013 in CALCULUS by homeworkhelp Mentor

2 Answers

+1 vote

The avarage value of f(x) in the interval [-1,3]

When

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f(2) = ?

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Note:image   Here image

Given that interval [-1,3]

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There fore

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Formula: image

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Formula: image

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Find f(2) = ?

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Formula: image

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answered Jan 19, 2013 by richardson Scholar

The average value of f (x ) in the interval [-1, 3], faverage  = 2.

0 votes

f '(x ) = 3x2 - 6x

Take integration to each side.

ʃ f '(x ) = ʃ (3x2 - 6x ) dx

f (x ) = ʃ (3x2 - 6x ) dx

= ʃ (3x2) dx - ʃ (6x ) dx

= 3ʃ x2 dx - 6ʃx dx

= 3(x3/3)  - 6(x2/2) + C

= x3  - 3x2 + C -> (1)

Condition is f(2) = 0.

0 = (2)3  - 3(2)2 + C

0 = 8  - 3(4) + C

0 = 8  - 12 + C

0 = -4 + C

C = 4

Substitute c value in(1)

f (x ) = x3  - 3x2 + 4

The average value of the function f (x ) over the interval [a, b] is image.

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Therefore the average value of f (x ) in the interval [-1, 3], faverage  = 2.

 

 

answered Jul 7, 2014 by joly Scholar

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