Question Number 84219 by Rio Michael last updated on 10/Mar/20
$$\int_{−\mathrm{1}} ^{\mathrm{1}} {e}^{\mid{x}\mid} {dx}\:=? \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
$$\int_{−\mathrm{1}} ^{\mathrm{0}} {e}^{−{x}} {dx}+\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}} {dx} \\ $$$$=\mid\frac{{e}^{−{x}} }{−\mathrm{1}}\mid_{−\mathrm{1}} ^{\mathrm{0}} +\mid\frac{{e}^{{x}} }{\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{{e}^{−\mathrm{0}} }{−\mathrm{1}}−\frac{{e}^{\mathrm{1}} }{−\mathrm{1}}+\frac{{e}}{\mathrm{1}}−\frac{{e}^{\mathrm{0}} }{\mathrm{1}} \\ $$$$=−\mathrm{1}+{e}+{e}−\mathrm{1}=\mathrm{2}\left({e}−\mathrm{1}\right) \\ $$