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Question
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`
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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