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November 26, 2014

November 26, 2014

Posted by **jimbo** on Monday, February 14, 2011 at 6:21pm.

i. g(x) = A-f(x) where A is a constant

ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx

iii. for the integral from 2 to 3 f(x)dx = -3A

a find the integral from 1 to 3 of f(x)dx in terms of A.

b. find the average value of g(x) in terms of A, over the interval [1,3]

c. Find the value of k if the integral from 0 to 1 f(x+1)dx=kA

need the steps for these problems worked out plz

- math -
**MathMate**, Monday, February 14, 2011 at 7:02pma.

Let

I1=∫f(x)dx from 1 to 2

I2=∫f(x)dx from 2 to 3

J1=∫g(x)dx from 1 to 2

J2=∫g(x)dx from 2 to 3

∫f(x)dx from 1 to 3

= I1 + I2

= J2 + I2 [I1=J2 from ii]

= ∫(f(x)+g(x))dx from 2 to 3

= ∫Adx from 2 to 3

= A

b.

The average value of g(x) from 1 to 3 is the integral of g(x) over the same limits divided by (3-1)=2.

∫g(x)dx 1 to 3

=∫(A-f(x))dx from 1 to 3 [from i]

=[∫Adx - ∫f(x)dx] from 1 to 3

=2A - A

=A

Average value of g(x) = A/(3-1)=A/2.

c.

∫f(x+1)dx from 0 to 1

= ∫f(x)dx from 1 to 2

= I1

= J2

= ∫(A-f(x))dx from 2 to 3

= A - (-3A) [ from ii ]

= 4A

Therefore if 4A=kA, k=4

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