# math

posted by on .

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