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 53212 by Joel578 last updated on 19/Jan/19

$$\mathrm{Let}f\left(x\right)=\frac{2x}{x^2+4}$$ $$$$ $$\left(a\right)\mathrm{Find}\underset{-b}{\stackrel{b}{\int }}f\left(x\right)dx,\mathrm{for}b>0$$ $$\left(b\right)\mathrm{Determine}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}f\left(x\right)dx\mathrm{is}\mathrm{convergent}\mathrm{or}\mathrm{not}$$

Commented byJoel578 last updated on 19/Jan/19

$$\left(a\right)f\left(x\right)=\frac{2x}{x^2+4}$$ $$f(-x)=\frac{2(-x)}{{(-x)}^2+4}=-\frac{2x}{x^2+4}=-f\left(x\right)$$ $$f\left(x\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}$$ $$\Rightarrow \underset{-b}{\stackrel{b}{\int }}f\left(x\right)dx=0$$

Commented byJoel578 last updated on 19/Jan/19

$$\mathrm{Please}\mathrm{help}\mathrm{with}\mathrm{part}\left(b\right)$$

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$1.\frac{2x}{x^2+4}isoddfunctionbecausef(-x)=-f\left(x\right)$$ $$2.theintregal{\int }_{-b}^b\frac{2x}{x^2+4}dx={\int }_{-b}^0\frac{2x}{x^2+4}dx+{\int }_0^b\frac{2x}{x^2+4}dx$$ $$I=I_1+I_2$$ $$though{I}_1=I_{2}but{I}_{1}representareaboundby$$ $$curve\frac{2x}{4+x^2}fromx=-btox=0butthatarea$$ $$isbelowxaxissonegetivearea.$$ $$I_{2}representtheareaboundbythecurve\frac{2x}{4+x^2}$$ $$fromx=0tox=babovexaxisso+vearea$$ $$hence{I}_1+I_2=0$$ $$thisismyopinionletotherscomment...$$ $$$$ $$$$

Commented byafachri last updated on 19/Jan/19

$$\underset{-b}{\stackrel{b}{\int }}\frac{2x}{x^2+4}dx=\underset{-b}{\stackrel{0}{\int }}\frac{2xdx}{x^2+4}+\underset{0}{\stackrel{b}{\int }}\frac{2xdx}{x^2+4}$$ $$=\ln (b^2+4)+\ln (b^2+4)$$ $$=2\ln (b^2+4)$$

Commented byafachri last updated on 19/Jan/19

Commented byafachri last updated on 19/Jan/19

$$\mathrm{in}\mathrm{my}\mathrm{opinion}\mathrm{to}\mathrm{the}\mathrm{part}\mathrm{B},\mathrm{we}\mathrm{can}\mathrm{define}$$ $$\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}f\left(x\right)\mathrm{is}\mathrm{convergent}\mathrm{or}\mathrm{not}\mathrm{by}\mathrm{check}\mathrm{the}$$ $$\mathrm{limit}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{divergent}\mathrm{while}\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=\mathrm{\infty }.$$ $$\underset{x\to \mathrm{\infty }}{\lim }\frac{2x}{x^2+4}=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2\left(\frac{2}{x}\right)}{x^2(1+\frac{4}{x^2})}$$ $$=\underset{x\to \mathrm{\infty }}{\lim }\frac{0}{1+0}$$ $$=0$$ $$f\left(x\right)=\frac{2x}{x^2+4}\mathrm{converges}\mathrm{so}\underset{-\mathrm{\infty }}{\stackrel{\mathrm{\infty }}{\int }}f\left(x\right)\mathrm{does}.$$

Commented byafachri last updated on 19/Jan/19

$$\mathrm{But}\mathrm{Mr}\mathrm{Tanmay},$$ $$\underset{-b}{\stackrel{0}{\int }}f\left(x\right)+\underset{0}{\stackrel{b}{\int }}=\mathrm{F}\left(x\right)\underset{-b}{\stackrel{0}{\mid }}+\mathrm{F}\left(x\right)\underset{0}{\stackrel{b}{\mid }}$$ $$=(\mathrm{F}\left(0\right)-\mathrm{F}(-b))+(\mathrm{F}\left(b\right)-\mathrm{F}\left(0\right))$$ $$=\mathrm{F}\left(b\right)+\mathrm{F}\left(b\right)$$ $$=2\mathrm{F}\left(b\right)$$ $$$$

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$intregalvalueissame...butfrom[graphone$$ $$areaisabovexaxisanotherbelowxsissooddfunction...$$

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

Commented byafachri last updated on 19/Jan/19

$$\mathrm{thank}\mathrm{you}\mathrm{Mr}\mathrm{Damian}\mathrm{and}\mathrm{all}\mathrm{of}\mathrm{you}\mathrm{Sir}$$ $$\mathrm{who}\mathrm{have}\mathrm{corrected}\mathrm{me}\mathrm{for}\mathrm{my}\mathrm{mistake}.$$ $$\mathrm{what}\mathrm{an}\mathrm{interesting}\mathrm{forum}\mathrm{here}.\mathrm{now}{\mathrm{i}}^{\prime }\mathrm{m}$$ $$\mathrm{understand}\mathrm{it}.$$

Commented byafachri last updated on 19/Jan/19

$$\mathrm{Yes}\mathrm{Mr}\mathrm{Tanmay}.\mathrm{but}\mathrm{what}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{questioning}$$ $$\mathrm{is}\mathrm{why}\underset{-b}{\stackrel{b}{\int }}\frac{2x}{x^2+4}=0?\mathrm{Well},\underset{-b}{\stackrel{b}{\int }}\frac{2x}{x^2+4}\mathrm{states}$$ $$\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{function}\mathrm{through}\mathrm{absis}.$$ $$\mathrm{i}\mathrm{hope}\mathrm{you}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{mind}\mathrm{for}\mathrm{my}\mathrm{situation}$$ $$\mathrm{Mr}\mathrm{Tanmay}.$$

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$wait...letassumeb=1000$$ $$thenfind{\int }_{-1000}^{1000}\frac{2x}{4+x^2}dxusinggraphapp..$$

Commented bytanmay.chaudhury50@gmail.com last updated on 19/Jan/19

Commented byafachri last updated on 19/Jan/19

$$\mathrm{Yes}\mathrm{Mr}\mathrm{Tanmay}.\mathrm{Done}.\mathrm{Thank}\mathrm{You}$$ $$\mathrm{so}\mathrm{much}.\mathrm{Really}\mathrm{appreciate}\mathrm{your}\mathrm{time}$$ $$\mathrm{Mr}\mathrm{Tanmay}$$

Commented byafachri last updated on 19/Jan/19

Commented byJDamian last updated on 19/Jan/19

$$itiswrong.f\left(x\right)isodd,butF\left(x\right)iseven.Then$$ $$F\left(b\right)=F(-b)\Rightarrow F\left(b\right)-F(-b)=F\left(b\right)-F\left(b\right)=0$$ $$$$ $$thiscommentwasintendedforan{afrachi}^{\prime }sanswer.$$

Commented byJoel578 last updated on 20/Jan/19

$$\mathrm{Thanks}\mathrm{you}\mathrm{Sir}\mathrm{afachri},\mathrm{tanmay},\mathrm{JDamian}\mathrm{for}\mathrm{the}\mathrm{answers}.$$ $$\mathrm{Actually}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{agree}\mathrm{with}\mathrm{Sir}\mathrm{tanmay}.{\mathrm{That}}^{\prime }\mathrm{s}$$ $$\mathrm{what}\mathrm{I}\mathrm{have}\mathrm{been}\mathrm{taught}\mathrm{in}\mathrm{Calculus}1.$$ $$\mathrm{There}\mathrm{is}\mathrm{a}\mathrm{difference}\mathrm{between}\mathrm{a}\mathrm{definite}\mathrm{integral}$$ $$\mathrm{and}\mathrm{a}\mathrm{definite}\mathrm{integral}\mathrm{as}\mathrm{an}\mathrm{area}$$

Commented bymaxmathsup by imad last updated on 23/Jan/19

$$foralloddfunctionintegrableon]-a,a[a\in \stackrel{-}{R}wehave$$ $${\int }_{-a}^{a}f\left(x\right)dx=0.$$

Answered by MJS last updated on 20/Jan/19

$$\left(\mathrm{b}\right)\mathrm{is}\mathrm{already}\mathrm{done}\mathrm{after}\left(\mathrm{a}\right)$$ $$\int \frac{2x}{x^2+4}dx=\ln (x^2+4)=F\left(x\right)$$ $$\Rightarrow g\left(b\right)=F\left(b\right)-F(-b)=0$$ $$\underset{b\to \mathrm{\infty }}{\lim }g\left(b\right)=\underset{b\to \mathrm{\infty }}{\lim 0}=0$$

Commented byJoel578 last updated on 20/Jan/19

$$thankyouverymuch$$

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19

$$b)itismyopinion...$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(x\right)dx$$ $$\underset{h\to -\mathrm{\infty }}{\lim }{\int }_h^{0}f\left(x\right)dx=-veA$$ $$\underset{k\to +\mathrm{\infty }}{\lim }{\int }_0^{k}f\left(x\right)dx=+veA$$ $$so{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2x}{4+x^2}dx=0$$ $$anootherway...$$ $$x=2tan\theta dx=2{sec}^2\theta d\theta$$ $${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{2\times 2tan\theta \times 2{sec}^2\theta d\theta }{4+4{tan}^2\theta }$$ $$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{8tan\theta {sec}^2\theta }{4{sec}^2\theta }d\theta$$ $$=2\mid lnsec\theta {\mid }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$$ $$=-2\mid lncos\theta {\mid }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$$ $$=-2[lncos\left(\frac{\pi }{2}\right)-lncos(-\frac{\pi }{2})]$$ $$=0$$ $$itismyopinion...letotherscomment...$$

Commented byJoel578 last updated on 20/Jan/19

$$thankyouverymuch$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com