Prove that int_a^bf(x)dx=f(a+b-x)dx - Mathematics and Statistics

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Prove that `int_a^bf(x)dx=f(a+b-x)dx.` Hence evaluate : `int_a^bf(x)/(f(x)+f(a-b-x))dx`

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Solution

`"Let "I = int_a^bf(x)dx`

Put x= a + b - t

∴ dx = -dt

When x = a, t = b and when x = b, t = a

`therefore I = int_b^af(a+b-t)(-dt)`

`therefore I = -int_b^af(a+b-t)dt`

`therefore I = int_a^bf(a+b-t)dt ... [because int_a^bf(x)dx=-int_b^af(x)dx]`

`therefore int_a^bf(x)dx = int_a^bf(a+b-x)dx ... [because int_a^bf(x)dx= int_a^bf(t)dt]`

`"Let "I = int_a^b(f(x))/(f(x)+f(a+b-x))dx ... (i)`

`therefore I = int_a^b(f(a+b-x))/(f(a+b-x)+f(a+b-(a+b-x)))dx`

`therefore I = int_a^b(f(a+b-x))/(f(a+b-x)+f(x))dx ... (ii)`

Adding (i) and (ii) we get

`2I = int_a^b(f(x)+f(a+b-x))/(f(x)+f(a+b-x))dx`

`therefore 2I = int_a^b1dx`

`therefore 2I = [x]_a^b`

`therefore I = (b-a)/2`

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2013-2014 (October)

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