If f (x) is of the form f (x) = a+b x+cx², show that
$integral subscript 0 superscript 1 straight f left parenthesis straight x right parenthesis space dx space equals space 1 over 6 open curly brackets straight f left parenthesis 0 right parenthesis space plus space 4 space straight f space open parentheses 1 half close parentheses plus straight f left parenthesis 1 right parenthesis close curly brackets.$

Solution

Here $straight f left parenthesis straight x right parenthesis space equals space straight a plus straight b plus cx squared$
$therefore space space straight f left parenthesis 0 right parenthesis space equals space straight a plus 0 plus 0 space equals space straight a comma space space space straight f open parentheses 1 half close parentheses space equals space straight a plus straight b over straight c plus straight c over 4 comma space space space straight f left parenthesis 1 right parenthesis space equals space straight a plus straight b plus straight c$
R.H.S. = $1 over 6 open curly brackets straight f left parenthesis 0 right parenthesis plus 4 space straight f open parentheses 1 half close parentheses plus straight f left parenthesis 1 right parenthesis close curly brackets space equals space 1 over 6 open curly brackets straight a plus 4 open parentheses straight a plus straight b over 2 plus straight c over 4 close parentheses plus left parenthesis straight a plus straight b plus straight c right parenthesis close curly brackets$
$equals 1 over 6 left curly bracket straight a plus 4 straight a plus 2 straight b plus straight c plus straight a plus straight b plus straight c right curly bracket space equals space 1 over 6 left parenthesis 6 straight a plus 3 straight b plus 2 straight c right parenthesis space equals space straight a plus straight b over 2 plus straight c over 3$
$straight L. straight H. straight S. space equals space integral subscript 0 superscript 1 straight f left parenthesis straight x right parenthesis space dx space equals space integral subscript 0 superscript 1 left parenthesis straight a plus bx plus cx squared right parenthesis space dx space equals space open square brackets ax plus bx squared over 2 plus cx cubed over 3 close square brackets subscript 0 superscript 1$
$equals space open parentheses straight a plus straight b over 2 plus straight c over 3 close parentheses space minus space left parenthesis 0 plus 0 plus 0 right parenthesis space equals space straight a plus straight b over 2 plus straight c over 3$
∴ L.H.S. = R.H.S. Hence the result.

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