JEE Main 20207 Jan, Morning shiftMathematicsMediumIntegral Calculus369× since 2002If 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:AOption A. ∫a−1b−1f(x+1)dx\int_{a - 1}^{b - 1} {f(x+1)dx}∫a−1b−1f(x+1)dxBOption B. ∫a+1b+1f(x+1)dx\int_{a + 1}^{b + 1} {f(x + 1)dx}∫a+1b+1f(x+1)dxCOption C. ∫a−1b−1f(x)dx\int_{a - 1}^{b - 1} {f(x)dx}∫a−1b−1f(x)dxDOption D. ∫a+1b+1f(x)dx\int_{a + 1}^{b + 1} {f(x)dx}∫a+1b+1f(x)dxCheckSkip · show solution
