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Grade: 12

                        

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. 

9 months ago

Answers : (1)

Aditya Gupta
2074 Points
							
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 :))
9 months ago
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