Question Number 78766 by M±th+et£s last updated on 20/Jan/20 | ||
$$\int\mathrm{2}\:{e}^{\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }} \:{dx} \\ $$ | ||
Answered by MJS last updated on 20/Jan/20 | ||
$$\mathrm{2}\int\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}\left({x}−\mathrm{2}\right)^{\mathrm{2}} }} {dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}−{x}\:\rightarrow\:{dx}=−{dt}\right] \\ $$$$=−\mathrm{2}\int\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} {dt}= \\ $$$$\:\:\:\:\:\mathrm{by}\:\mathrm{parts}: \\ $$$$\:\:\:\:\:{u}'=\mathrm{1}\:\rightarrow\:{u}={t} \\ $$$$\:\:\:\:\:{v}=\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} \:\rightarrow\:{v}'=−\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} }{{t}^{\mathrm{3}} } \\ $$$$=−\mathrm{2}{t}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} +\mathrm{2}\int−\frac{\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} }{{t}^{\mathrm{2}} }{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}{t}}\:\rightarrow\:{dt}=−\sqrt{\mathrm{2}}{t}^{\mathrm{2}} {du}\right] \\ $$$$=−\mathrm{2}{t}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} +\sqrt{\mathrm{2}\pi}\int\frac{\mathrm{2e}^{{u}^{\mathrm{2}} } }{\sqrt{\pi}}{du}= \\ $$$$=−\mathrm{2}{t}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} +\sqrt{\mathrm{2}\pi}\mathrm{erfi}\:{u}\:= \\ $$$$=−\mathrm{2}{t}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{t}^{\mathrm{2}} }} +\sqrt{\mathrm{2}\pi}\mathrm{erfi}\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}{t}}= \\ $$$$=−\mathrm{2}\left(\mathrm{2}−{x}\right)\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−{x}\right)^{\mathrm{2}} }} +\sqrt{\mathrm{2}\pi}\mathrm{erfi}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}\left(\mathrm{2}−{x}\right)}\:+{C} \\ $$ | ||
Commented by mind is power last updated on 20/Jan/20 | ||
$$\mathrm{Nice} \\ $$ | ||
Commented by M±th+et£s last updated on 20/Jan/20 | ||
$${thank}\:{you}\:{sir} \\ $$ | ||