Aditya Gupta
Last Activity: 4 Years ago
this is a rather easy problem.
we simply need to apply the property: ∫a to b g(x)dx = ∫a to b g(a+b – x)dx
now, let J= (1/(a+b)) * ∫a to b x(f(x) + f(x+1))dx.......(1)
so J= (1/(a+b)) * ∫a to b (a+b – x)(f(a+b – x) + f(a+b – x+1))dx........(2)
now, since f(a+b+1 – x)= f(x) for all x, replace x by x+1 to get f(a+b – x)= f(x+1)
so, (2) becomes J= (1/(a+b)) * ∫a to b (a+b – x)(f(x+1) + f(x))dx......(3)
add (1) and (3)
2J= (1/(a+b)) * ∫a to b (a+b)(f(x+1) + f(x))dx
or 2J= ∫a to b (f(x+1) + f(x))dx
or 2J= ∫a to b f(x)dx + ∫a to b f(x+1)dx
now, ∫a to b f(x+1)dx= ∫a to b f(a+b – x+1)dx= ∫a to b f(x)dx
so, 2J= ∫a to b f(x)dx + ∫a to b f(x)dx
or J= ∫a to b f(x)dx
let x= y+1
then J= ∫a – 1 to b – 1 f(y+1)dy
since y is a dummy variable, replace it back by x.
so J= ∫a – 1 to b – 1 f(x+1)dx
option (3)
KINDLY APPROVE :))
