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Kindly show me the solution for the given sum. It is taken from the January session of the JEE main examinations.

 Kindly show me the solution for the given sum. It is taken from the January session of the JEE main examinations. 

Question Image
Grade:12

1 Answers

Aditya Gupta
2080 Points
one year 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 :))

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