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Question Number 6700 by jose14918645@gmail.com last updated on 14/Jul/16

∫(e^(−2+2) xdx

$$\int\left({e}^{−\mathrm{2}+\mathrm{2}} {xdx}\right. \\ $$

Answered by FilupSmith last updated on 15/Jul/16

=∫e^0 xdx  =∫xdx  =(1/2)x^2 +c

$$=\int{e}^{\mathrm{0}} {xdx} \\ $$$$=\int{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{c} \\ $$

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