Evaluate the following definite integral as limit of a sum.
$integral subscript 0 superscript 5 left parenthesis straight x minus 1 right parenthesis dx$

Solution

Comparing $integral subscript 0 superscript 5 left parenthesis straight x minus 1 right parenthesis dx space with space integral subscript straight a superscript straight b space straight f left parenthesis straight x right parenthesis space dx comma space space space we space get comma$
f(x) = x - 1, a = 0, b = 5
$straight f left parenthesis straight a plus 2 space straight h right parenthesis space equals space straight f left parenthesis 2 space straight h right parenthesis space equals space 2 space straight h space minus space 1 comma space space..... comma space space left parenthesis straight a plus stack straight n minus 1 with bar on top straight h right parenthesis space equals space straight f left parenthesis stack straight n minus 1 with bar on top straight h right parenthesis space equals space left parenthesis straight n minus 1 right parenthesis space straight h space minus 1$
Now $integral subscript straight a superscript straight b straight f left parenthesis straight x right parenthesis space equals space Lt with straight h rightwards arrow 0 below space straight h open square brackets straight f left parenthesis straight a right parenthesis plus straight f left parenthesis straight a plus straight h right parenthesis plus straight f left parenthesis straight a plus 2 straight h right parenthesis plus... plus straight f left parenthesis straight a plus stack straight n minus 1 with bar on top straight h right parenthesis close square brackets$
$equals space Lt with straight h rightwards arrow 0 below space straight h open square brackets left parenthesis negative 1 right parenthesis plus left parenthesis straight h minus 1 right parenthesis plus left parenthesis 2 straight h minus 1 right parenthesis plus.... plus open curly brackets left parenthesis straight n minus 1 right parenthesis straight h minus 1 close curly brackets close square brackets equals space Lt with straight h space rightwards arrow 0 below space straight h open square brackets negative straight n plus straight h open curly brackets 1 plus 2 plus 3 plus... plus left parenthesis straight n minus 1 right parenthesis close curly brackets close square brackets space equals space Lt with straight h rightwards arrow 0 below space straight h open square brackets negative straight n plus straight h fraction numerator left parenthesis straight n minus 1 minus straight n right parenthesis over denominator 2 end fraction close square brackets$
$equals space Lt with straight h rightwards arrow 0 below open square brackets negative nh plus fraction numerator left parenthesis straight n space straight h right parenthesis space left parenthesis nh space minus space straight h right parenthesis over denominator 2 end fraction close square brackets space equals space Lt with straight h rightwards arrow 0 below open square brackets negative left parenthesis 5 minus 0 right parenthesis plus left parenthesis 5 minus 0 right parenthesis fraction numerator left parenthesis 5 minus 0 minus straight h right parenthesis over denominator 2 end fraction close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space straight n space straight h space equals space straight b space minus straight a space equals space 5 space minus space 0 close square brackets equals space minus 5 plus fraction numerator 5 left parenthesis 5 minus 0 right parenthesis over denominator 2 end fraction equals negative 5 plus 25 over 2 equals fraction numerator negative 10 plus 25 over denominator 2 end fraction equals 15 over 2$

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Integrals

Evaluate the following definite integral as limit of a sum.
$integral subscript 0 superscript 5 left parenthesis straight x minus 1 right parenthesis dx$

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