Evaluate $integral subscript 1 superscript 4 left parenthesis straight x squared minus straight x right parenthesis space dx$ as the limit of a sum.

Solution

Comparing $integral subscript 1 superscript 4 left parenthesis straight x squared minus straight x right parenthesis dx space space with space integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx comma space we space get$
$straight f left parenthesis straight x right parenthesis space equals space straight x squared minus straight x comma space space straight a space equals space 1 comma space space space space straight b space equals space 4$
$space space space space space space straight f left parenthesis straight a right parenthesis space equals space straight f left parenthesis 1 right parenthesis space equals space 1 minus 1 space equals space 0 straight f left parenthesis straight a plus straight h right parenthesis space equals space straight f left parenthesis 1 plus straight h right parenthesis space equals space left parenthesis 1 plus straight h right parenthesis squared minus left parenthesis 1 plus straight h right parenthesis straight f left parenthesis straight a plus 2 straight h right parenthesis space equals space straight f left parenthesis 1 plus 2 straight h right parenthesis space equals space left parenthesis 1 plus 2 straight h right parenthesis squared minus left parenthesis 1 plus 2 straight h right parenthesis$
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$straight f left parenthesis straight a plus stack straight n minus 1 with bar on top space straight h right parenthesis space equals space straight f left parenthesis 1 plus stack straight n minus 1 with bar on top space straight h right parenthesis space equals space left parenthesis 1 plus stack straight n minus 1 with bar on top right parenthesis squared minus left parenthesis 1 plus stack straight n minus 1 with bar on top space straight h right parenthesis$
Now $integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space dx space equals space Lt with straight h rightwards arrow 0 below space straight h space left square bracket straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus..... plus straight f left parenthesis straight a plus stack straight n minus 1 with bar on top space straight h right parenthesis right square bracket$
$therefore space space space integral subscript 1 superscript 4 left parenthesis straight x squared minus straight x right parenthesis dx space equals space Lt with straight h rightwards arrow 0 below space straight h left square bracket 0 plus open curly brackets left parenthesis 1 plus straight h right parenthesis squared minus left parenthesis 1 plus straight h right parenthesis close curly brackets space plus space open curly brackets left parenthesis 1 plus 2 straight h right parenthesis squared minus left parenthesis 1 plus 2 straight h right parenthesis close curly brackets plus.....$
$plus open curly brackets left parenthesis 1 plus stack straight n minus 1 with bar on top space straight h right parenthesis squared minus left parenthesis 1 plus stack straight n minus 1 with bar on top space straight h right parenthesis close curly brackets right square bracket$
$equals space Lt with straight h rightwards arrow 0 below space straight h left square bracket open curly brackets straight n plus 2 straight h space left parenthesis 1 plus 2 plus 3 plus... plus stack straight n minus 1 with bar on top right parenthesis plus straight h squared left parenthesis 1 squared plus 2 squared plus.... plus left parenthesis straight n minus 1 right parenthesis squared close curly brackets$
$negative open curly brackets straight n plus left parenthesis 1 plus 2 plus.... plus stack straight n minus 1 with bar on top right parenthesis space straight h close curly brackets right square bracket$

   $equals space Lt with straight h rightwards arrow 0 below space straight h open square brackets straight n plus 2 straight h plus fraction numerator left parenthesis straight n minus 1 right parenthesis straight n over denominator 2 end fraction plus straight h squared fraction numerator left parenthesis straight n minus 1 right parenthesis space left parenthesis straight n right parenthesis space left parenthesis 2 straight n minus 1 right parenthesis over denominator 6 end fraction minus straight n minus fraction numerator left parenthesis straight n minus 1 right parenthesis space left parenthesis straight n right parenthesis over denominator 2 end fraction straight h close square brackets equals space Lt with straight h rightwards arrow 0 below space open square brackets straight n space straight h space plus space left parenthesis straight n space straight h space minus space straight h right parenthesis space left parenthesis nh right parenthesis space plus space fraction numerator left parenthesis nh minus straight h right parenthesis space left parenthesis nh right parenthesis space left parenthesis 2 nh minus straight h right parenthesis over denominator 6 end fraction minus nh minus fraction numerator left parenthesis nh minus straight h right parenthesis space left parenthesis nh right parenthesis over denominator 2 end fraction close square brackets equals space Lt with straight h rightwards arrow 0 below space open square brackets 3 plus left parenthesis 3 minus straight h right parenthesis space left parenthesis 3 right parenthesis plus space fraction numerator left parenthesis 3 minus straight h right parenthesis space left parenthesis 6 minus straight h right parenthesis over denominator 6 end fraction minus 3 minus fraction numerator left parenthesis 3 minus straight h right parenthesis space left parenthesis 3 right parenthesis over denominator 2 end fraction close square brackets$
   $open square brackets because space space straight n space straight h space equals space straight b space minus straight a space equals space 4 space minus space 1 space equals space 3 close square brackets$
   $space equals 3 plus left parenthesis 3 minus 0 right parenthesis space left parenthesis 3 right parenthesis space plus fraction numerator left parenthesis 3 minus 0 right parenthesis space left parenthesis 3 right parenthesis space left parenthesis 6 minus 0 right parenthesis over denominator 6 end fraction minus 3 minus fraction numerator left parenthesis 3 minus 0 right parenthesis space left parenthesis 3 right parenthesis over denominator 2 end fraction space equals space 3 plus 9 plus 9 minus 3 minus 9 over 2 space equals space 27 over 2$

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Integrals

Evaluate $integral subscript 1 superscript 4 left parenthesis straight x squared minus straight x right parenthesis space dx$ as the limit of a sum.

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