Let f:R→R be a continuous function such that f(x)+f(x+1)=2, for all x∈R. If I1=∫80f(x)dx and I2=∫3−1f(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=4∫20f(x)dxI2=∫3−1f(x)dx=∫40f(x+1)dx=∫40(2−f(x))dxI2=8−2∫20f(x)dxI1+2I2=16