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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.

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6 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 :))
```
6 months ago
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