Posted by jimbo on Monday, February 14, 2011 at 6:21pm.
LEt f and g be continous functions with the following properties
i. g(x) = Af(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 7:02pm
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 (31)=2.
∫g(x)dx 1 to 3
=∫(Af(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/(31)=A/2.
c.
∫f(x+1)dx from 0 to 1
= ∫f(x)dx from 1 to 2
= I1
= J2
= ∫(Af(x))dx from 2 to 3
= A  (3A) [ from ii ]
= 4A
Therefore if 4A=kA, k=4
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