Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 32361 by prof Abdo imad last updated on 23/Mar/18

let give a>0 find  ∫_0 ^∞    (e^(−x) /(√(x+a))) dx.

$${let}\:{give}\:{a}>\mathrm{0}\:{find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}} }{\sqrt{{x}+{a}}}\:{dx}. \\ $$

Commented byabdo imad last updated on 25/Mar/18

the ch.(√(x+a)) =t  give x+a =t^2  ⇒ x=t^2 −a  ∫_0 ^∞   (e^(−x) /(√(x+a))) dx = ∫_(√a) ^(+∞)    (e^(−(t^2 −a)) /t) 2t dt  =2 e^a   ∫_(√a) ^(+∞)  e^(−t^2 )  dt =2 e^a ( ∫_0 ^∞  e^(−t^2 )  −∫_0 ^(√a)  e^(−t^2 ) dt)  =2 e^a  ((√π)/2)  −2 e^a   ∫_0 ^(√a)   e^(−t^2 ) dt  =e^a (√π)  −2 e^a  ∫_0 ^(√a)   e^(−t^2 )  dt  .

$${the}\:{ch}.\sqrt{{x}+{a}}\:={t}\:\:{give}\:{x}+{a}\:={t}^{\mathrm{2}} \:\Rightarrow\:{x}={t}^{\mathrm{2}} −{a} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} }{\sqrt{{x}+{a}}}\:{dx}\:=\:\int_{\sqrt{{a}}} ^{+\infty} \:\:\:\frac{{e}^{−\left({t}^{\mathrm{2}} −{a}\right)} }{{t}}\:\mathrm{2}{t}\:{dt} \\ $$ $$=\mathrm{2}\:{e}^{{a}} \:\:\int_{\sqrt{{a}}} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } \:{dt}\:=\mathrm{2}\:{e}^{{a}} \left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } \:−\int_{\mathrm{0}} ^{\sqrt{{a}}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\right) \\ $$ $$=\mathrm{2}\:{e}^{{a}} \:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:−\mathrm{2}\:{e}^{{a}} \:\:\int_{\mathrm{0}} ^{\sqrt{{a}}} \:\:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$ $$={e}^{{a}} \sqrt{\pi}\:\:−\mathrm{2}\:{e}^{{a}} \:\int_{\mathrm{0}} ^{\sqrt{{a}}} \:\:{e}^{−{t}^{\mathrm{2}} } \:{dt}\:\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com