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 20873 by Joel577 last updated on 05/Sep/17

∫_1 ^5 (e^x /x^2 ) dx

$$\int_{\mathrm{1}} ^{\mathrm{5}} \frac{{e}^{{x}} }{{x}^{\mathrm{2}} }\:{dx} \\ $$

Answered by alex041103 last updated on 08/Sep/17

First we apply integration by parts  ∫_( a) ^b udv=uv∣_a ^b  − ∫_( a) ^b vdu  u=e^x           dv=x^(−2) dx  du=e^x dx  v=−x^(−1)     ∫_( 1) ^5 (e^x /x^2 )dx=(e^x /x) ∣_5 ^1  + ∫^( 5) _( 1) (e^x /x)dx  =e−(e^5 /5)+∫_( 1) ^5 (e^x /x)dx  Now it′s time to say that the ∫(e^x /x)dx  is not elementary and we define  Ei(x)+C=∫(e^x /x)dx  Now we use e^x =Σ_(n=0) ^∞ (x^n /(n!))  And we can derive that  Ei(x)=Σ_(n=0) ^∞ (x^n /(n∙n!))  Using that we can approximate the integral  ∫_( 1) ^5 (e^x /x^2 )dx=e−(e^5 /5)+Ei(5)−Ei(1)≈11.3258  Note: In order for the series expansion  of Ei(x) to converge we use C=0

$${First}\:{we}\:{apply}\:{integration}\:{by}\:{parts} \\ $$$$\underset{\:{a}} {\overset{{b}} {\int}}{udv}={uv}\mid_{{a}} ^{{b}} \:−\:\underset{\:{a}} {\overset{{b}} {\int}}{vdu} \\ $$$${u}={e}^{{x}} \:\:\:\:\:\:\:\:\:\:{dv}={x}^{−\mathrm{2}} {dx} \\ $$$${du}={e}^{{x}} {dx}\:\:{v}=−{x}^{−\mathrm{1}} \\ $$$$ \\ $$$$\underset{\:\mathrm{1}} {\overset{\mathrm{5}} {\int}}\frac{{e}^{{x}} }{{x}^{\mathrm{2}} }{dx}=\frac{{e}^{{x}} }{{x}}\:\mid_{\mathrm{5}} ^{\mathrm{1}} \:+\:\underset{\:\mathrm{1}} {\int}^{\:\mathrm{5}} \frac{{e}^{{x}} }{{x}}{dx} \\ $$$$={e}−\frac{{e}^{\mathrm{5}} }{\mathrm{5}}+\underset{\:\mathrm{1}} {\overset{\mathrm{5}} {\int}}\frac{{e}^{{x}} }{{x}}{dx} \\ $$$${Now}\:{it}'{s}\:{time}\:{to}\:{say}\:{that}\:{the}\:\int\frac{{e}^{{x}} }{{x}}{dx} \\ $$$${is}\:{not}\:{elementary}\:{and}\:{we}\:{define} \\ $$$${Ei}\left({x}\right)+{C}=\int\frac{{e}^{{x}} }{{x}}{dx} \\ $$$${Now}\:{we}\:{use}\:{e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!} \\ $$$${And}\:{we}\:{can}\:{derive}\:{that} \\ $$$${Ei}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}\centerdot{n}!} \\ $$$${Using}\:{that}\:{we}\:{can}\:{approximate}\:{the}\:{integral} \\ $$$$\underset{\:\mathrm{1}} {\overset{\mathrm{5}} {\int}}\frac{{e}^{{x}} }{{x}^{\mathrm{2}} }{dx}={e}−\frac{{e}^{\mathrm{5}} }{\mathrm{5}}+{Ei}\left(\mathrm{5}\right)−{Ei}\left(\mathrm{1}\right)\approx\mathrm{11}.\mathrm{3258} \\ $$$${Note}:\:{In}\:{order}\:{for}\:{the}\:{series}\:{expansion} \\ $$$${of}\:{Ei}\left({x}\right)\:{to}\:{converge}\:{we}\:{use}\:{C}=\mathrm{0} \\ $$

Commented by Joel577 last updated on 09/Sep/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com