If f(a+b+1−x)=f(x)∀x, where a and b are fixed positive real numbers, then 1(a+b)b∫ax(f(x)+f(x+1))dx is equal to
A
b−1∫a−1f(x)dx
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B
b+1∫a+1f(x+1)dx
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C
b−1∫a−1f(x+1)dx
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D
b+1∫a+1f(x)dx
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Solution
The correct option is Cb−1∫a−1f(x+1)dx f(a+b+1−x)=f(x)...(1) x→x+1 f(a+b−x)=f(x+1)...(2) I=1a+bb∫ax(f(x)+f(x+1))dx...(3)
From (1) and (2) I=1a+bb∫a(a+b−x)(f(x+1)+f(x))dx...(4)
Adding (3) and (4) 2I=b∫a(f(x)+f(x+1))dx⇒2I=b∫af(x+1)dx+b∫af(x)dx⇒2I=b∫af(a+b−x+1)dx+b∫af(x)dx⇒2I=2b∫af(x)dx⇒I=b∫af(x)dx
Let x=t+1,dx=dt ⇒I=b−1∫a−1f(t+1)dt⇒I=b−1∫a−1f(x+1)dx