Question

If $straight f left parenthesis straight a plus straight b minus straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis comma space space then space integral subscript straight a superscript straight b straight x space straight f left parenthesis straight x right parenthesis space dx$ is equal to

$fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight b minus straight x right parenthesis space dx$
$fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight b plus straight x right parenthesis space dx$
$fraction numerator straight b minus straight a over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx$
$fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b space straight f left parenthesis straight x right parenthesis space dx$

Solution

fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b space straight f left parenthesis straight x right parenthesis space dx

Let I = $integral subscript straight a superscript straight b space straight x space straight f left parenthesis straight x right parenthesis space dx$ ...(1)
$therefore space space space space straight I space equals space integral subscript straight a superscript straight b left parenthesis straight a plus straight b minus straight x right parenthesis space straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx space space space space space space space space space space space space space open square brackets because space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b straight f left parenthesis straight a plus straight b minus straight x right parenthesis space dx close square brackets therefore space space space space straight I space equals space integral subscript straight a superscript straight b left parenthesis straight a plus straight b minus straight x right parenthesis space dx space space space space space space space space space... left parenthesis 2 right parenthesis$
$left square bracket because space space straight f left parenthesis straight a plus straight b minus straight x right parenthesis space equals space straight f left parenthesis straight x right parenthesis right square bracket$
Adding (1) and (2), we get,
$2 space straight I space equals space integral subscript straight a superscript straight b left parenthesis straight x plus straight a plus straight b minus straight x right parenthesis space straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript straight a superscript straight b left parenthesis straight a plus straight b right parenthesis space straight f left parenthesis straight x right parenthesis space dx$
$therefore space space space 2 space space straight I space equals space left parenthesis straight a plus straight b right parenthesis space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space space space rightwards double arrow space space space space straight I space equals space fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx therefore space space space integral subscript straight a superscript straight b space straight x space straight f left parenthesis straight x right parenthesis space dx space equals space fraction numerator straight a plus straight b over denominator 2 end fraction integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx$
$therefore space space space left parenthesis straight D right parenthesis space is space correct space answer.$

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