1g'(x) = f(x)
g'(x+λ)=f(x+λ)=f(x)=g'(x)
≡ g(x+λ)=g(x) + c
since g(0) = 0
now we are left to prove 0∫λf(x)dx = 0
let I =0∫λf(x)dx x→ x - λ/2
I = -λ/2∫λ/2 -f(x)dx
=> 2I = 0.
f(x) is a periodic fn with pd λ and f(x-\frac{\lambda }{2})=-f(x),then show thta g(x+\lambda)=g(x) where g(x)=\int_{0}^{x}{f(t)dt}
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3 Answers
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24my soln a bit diff ...though i am sure many will say its one and same thing[3][3]
we write g(x+λ)=( 0 to x+λ)∫f(t)dt
=> g(x+λ)=(0 to x)∫f(t)dt+ (x to x+λ)∫f(t)dt
=> g(x+λ)=g(x)+ (x to x+λ)∫f(t)dt
=> g(x+λ) =g(x)+(0 to λ/2)∫f(t)dt + (λ/2 to λ)∫f(t)dt
Now ,
(0 to λ/2)∫f(t)dt
put u - λ/2=t => du=dt
=> (λ/2 to λ)∫f(u- λ/2) du
=> -[(λ/2 to λ)∫f(t) du] => (0 to λ/2)∫f(t)dt=-[(λ/2 to λ)∫f(t) dt]
putting this back in eqn
g(x+λ)=g(x)-[(λ/2 to λ)∫f(t) dt]+(λ/2 to λ)∫f(t)dt
=> g(x+λ) =g(x)
** Sorry for all this mess ..but latex is not opening on my PC[2]