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
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
Average value of g(x) = A/(3-1)=A/2.
∫f(x+1)dx from 0 to 1
= ∫f(x)dx from 1 to 2
= ∫(A-f(x))dx from 2 to 3
= A - (-3A) [ from ii ]
Therefore if 4A=kA, k=4
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