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Question

Let f:RR be a continuous function such that f(x)+f(x+1)=2, for all xR. If I1=80f(x)dx and I2=31f(x)dx then the value of I1+2I2 is equal to

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Solution

f(x)+f(x+1)=2(1)x(x+1)f(x+1)+f(x+2)=2(ii)by (i) & (ii)f(x)f(x+2)=0f(x+2)=f(x)
f(x) is periodic with T=2
I1=2×40f(x)dx=420f(x)dxI2=31f(x)dx=40f(x+1)dx=40(2f(x))dxI2=8220f(x)dxI1+2I2=16

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