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LEt f and g be continous functions with the following properties

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 -

    a.
    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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