Integration Questions

Question Number 94278 by  M±th+et+s last updated on 17/May/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{\sqrt{{x}}} }{\sqrt{{x}}}{dx} \\$$

Commented by PRITHWISH SEN 2 last updated on 17/May/20

$$\mathrm{e}^{\sqrt{\mathrm{x}}} =\mathrm{t} \\$$$$\mathrm{then}\:\:\:\:\frac{\mathrm{e}^{\sqrt{\mathrm{x}}} }{\mathrm{2}\sqrt{\mathrm{x}}}\:\mathrm{dx}\:=\:\mathrm{dt} \\$$$$\therefore\:\int\mathrm{2dt}=\mathrm{2}\:\mathrm{t}+\mathrm{C}\:=\mathrm{2}\:\mathrm{e}^{\sqrt{\mathrm{x}}} +\mathrm{C}\:\:\:\mathrm{C}=\:\mathrm{constant} \\$$

Commented by  M±th+et+s last updated on 17/May/20

$${thanks}\:{sir}\:{but}\:{its}\:\int_{\mathrm{0}} ^{\mathrm{1}} \\$$

Commented by mathmax by abdo last updated on 17/May/20

$${changement}\:\sqrt{{x}}={t}\:{give}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{e}^{\sqrt{{x}}} }{\sqrt{{x}}}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{e}^{{t}} }{{t}}\left(\mathrm{2}{t}\right){dt} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{{t}} \:{dt}\:=\mathrm{2}\left[{e}^{{t}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=\mathrm{2}\left({e}−\mathrm{1}\right)\:=\mathrm{2}{e}−\mathrm{2} \\$$

Commented by  M±th+et+s last updated on 17/May/20

$${thank}\:{you}\:{sir} \\$$

Commented by mathmax by abdo last updated on 17/May/20

$${you}\:{are}\:{welcome}. \\$$