MathematicsMedium210×since 2002Q3387If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then 1a+b∫abx(f(x)+f(x+1))dx{1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dxa+b1∫abx(f(x)+f(x+1))dx is equal to:A∫a−1b−1f(x+1)dx\int_{a - 1}^{b - 1} {f(x+1)dx}∫a−1b−1f(x+1)dxB∫a+1b+1f(x+1)dx\int_{a + 1}^{b + 1} {f(x + 1)dx}∫a+1b+1f(x+1)dxC∫a−1b−1f(x)dx\int_{a - 1}^{b - 1} {f(x)dx}∫a−1b−1f(x)dxD∫a+1b+1f(x)dx\int_{a + 1}^{b + 1} {f(x)dx}∫a+1b+1f(x)dxCheck answerSkip