Question Number 150247 by mathdanisur last updated on 10/Aug/21 | ||
$$\underset{\:\mathrm{2}} {\overset{\:\mathrm{6}} {\int}}\:\left(\mathrm{x}-\mathrm{1}\right)\left(\mathrm{x}-\mathrm{2}\right)\left(\mathrm{x}-\mathrm{3}\right)...\left(\mathrm{x}-\mathrm{9}\right)\:\mathrm{dx}\:=\:? \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{1}\left.\:\left.\:\:\:\:\mathrm{b}\right)\mathrm{0}\:\:\:\:\:\mathrm{c}\right)\mathrm{6}!\:\:\:\:\:\mathrm{d}\right)-\mathrm{2}\:\:\:\:\mathrm{e}\right)\mathrm{4}! \\ $$ | ||
Commented by amin96 last updated on 10/Aug/21 | ||
$$\int_{\mathrm{2}} ^{\mathrm{6}} \underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\prod}}\left({x}−{n}\right){dx}=\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\prod}}\int_{\mathrm{2}} ^{\mathrm{6}} \left({x}−{n}\right){dx}= \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\prod}}\left(\mid_{\mathrm{2}} ^{\mathrm{6}} \left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{nx}\right)\right)=\underset{{n}=\mathrm{1}} {\overset{\mathrm{9}} {\prod}}\left(\mathrm{16}−\mathrm{4}{n}\right)=\mathrm{12}\centerdot\mathrm{8}\centerdot\mathrm{4}\centerdot\mathrm{0}....=\mathrm{0} \\ $$ | ||
Commented by mr W last updated on 10/Aug/21 | ||
$${maybe}\:{the}\:{question}\:{is} \\ $$$$\underset{\:\mathrm{2}} {\overset{\:\mathrm{8}} {\int}}\:\left(\mathrm{x}-\mathrm{1}\right)\left(\mathrm{x}-\mathrm{2}\right)\left(\mathrm{x}-\mathrm{3}\right)...\left(\mathrm{x}-\mathrm{9}\right)\:\mathrm{dx}\:=\:? \\ $$ | ||
Commented by amin96 last updated on 10/Aug/21 | ||
$$\left.{Thanks}\right) \\ $$ | ||
Commented by mathdanisur last updated on 10/Aug/21 | ||
$${Thank}\:{you}\:{ser}\:{cool} \\ $$ | ||
Commented by mr W last updated on 10/Aug/21 | ||
$${obviously}\:{wrong}! \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){g}\left({x}\right){dx}\neq\left(\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\right)×\left(\int_{{a}} ^{{b}} {g}\left({x}\right){dx}\right) \\ $$ | ||
Commented by mr W last updated on 10/Aug/21 | ||
$${all}\:{answers}\:{given}\:{are}\:{wrong}! \\ $$ | ||
Commented by mr W last updated on 10/Aug/21 | ||
$${just}\:{expand}\:{and}\:{you}'{ll}\:{get}\:{the}\:{right} \\ $$$${answer}\:−\frac{\mathrm{1664}}{\mathrm{5}}. \\ $$ | ||
Commented by mathdanisur last updated on 10/Aug/21 | ||
$$\mathrm{Thankyou}\:\mathrm{Ser},\:\mathrm{so}\:\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{now}.? \\ $$ | ||
Commented by mr W last updated on 10/Aug/21 | ||
$$\underset{\:\mathrm{2}} {\overset{\:\mathrm{8}} {\int}}\:\left(\mathrm{x}-\mathrm{1}\right)\left(\mathrm{x}-\mathrm{2}\right)\left(\mathrm{x}-\mathrm{3}\right)...\left(\mathrm{x}-\mathrm{9}\right)\:\mathrm{dx}\:=\:\mathrm{0} \\ $$ | ||
Commented by mr W last updated on 10/Aug/21 | ||
$$\underset{\:\mathrm{5}−{a}} {\overset{\:\mathrm{5}+{a}} {\int}}\:\left(\mathrm{x}-\mathrm{1}\right)\left(\mathrm{x}-\mathrm{2}\right)\left(\mathrm{x}-\mathrm{3}\right)...\left(\mathrm{x}-\mathrm{9}\right)\:\mathrm{dx}\:=\:\mathrm{0} \\ $$$${a}\in\mathbb{R} \\ $$ | ||
Commented by mathdanisur last updated on 10/Aug/21 | ||
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$ | ||