If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then ${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ is equal to:

Question

If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then

${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ is equal to:

$\int_{a - 1}^{b - 1} {f(x+1)dx} $

$\int_{a + 1}^{b + 1} {f(x + 1)dx} $

$\int_{a - 1}^{b - 1} {f(x)dx} $

$\int_{a + 1}^{b + 1} {f(x)dx} $

Solution

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