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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.

```
9 months 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)dxnow, 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))dxor 2J= ∫a to b (f(x+1) + f(x))dxor 2J= ∫a to b f(x)dx + ∫a to b f(x+1)dxnow, ∫a to b f(x+1)dx= ∫a to b f(a+b – x+1)dx= ∫a to b f(x)dxso, 2J= ∫a to b f(x)dx + ∫a to b f(x)dxor J= ∫a to b f(x)dxlet x= y+1then J= ∫a – 1 to b – 1 f(y+1)dysince y is a dummy variable, replace it back by x.so J= ∫a – 1 to b – 1 f(x+1)dxoption (3)KINDLY APPROVE :))
```
9 months ago
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