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 42358 by Tawa1 last updated on 24/Aug/18

∫_( −1) ^( 1)  (x^(2015) /(((1 + x))^(1/(2015))   +  ((1 − x))^(1/(2015))  ))  dx

$$\int_{\:−\mathrm{1}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2015}} }{\sqrt[{\mathrm{2015}}]{\mathrm{1}\:+\:\mathrm{x}}\:\:+\:\:\sqrt[{\mathrm{2015}}]{\mathrm{1}\:−\:\mathrm{x}}\:}\:\:\mathrm{dx} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

Υ(x)=(x^(2015) /((1+x)^(1/(2015)) +(1−x)^(1/(2015)) ))  Υ(−x)=(((−x)^(2015) )/((1−x)^(1/(2015)) +(1+x)^(1/(2015)) ))=−Υ(x)  so f(x) is odd function...  ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx formula...  ∫_(−1) ^1 Υ(x)dx=∫_(−1) ^1 Υ(1−1−x)dx                           =∫_(−1) ^1 −Υ(x)dx  so 2∫_(−1) ^1 Υ(x)dx=0    ∫_(−1) ^1 Υ(x)dx=0  hence ∫_(−1) ^1  (x^(2015) /((1+x)^(1/(2015)) +(1−x)^(1/(2015)) ))dx    =0

$$\Upsilon\left({x}\right)=\frac{{x}^{\mathrm{2015}} }{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{2015}}} +\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2015}}} } \\ $$$$\Upsilon\left(−{x}\right)=\frac{\left(−{x}\right)^{\mathrm{2015}} }{\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2015}}} +\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{2015}}} }=−\Upsilon\left({x}\right) \\ $$$${so}\:{f}\left({x}\right)\:{is}\:{odd}\:{function}... \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx}\:{formula}... \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \Upsilon\left({x}\right){dx}=\int_{−\mathrm{1}} ^{\mathrm{1}} \Upsilon\left(\mathrm{1}−\mathrm{1}−{x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int_{−\mathrm{1}} ^{\mathrm{1}} −\Upsilon\left({x}\right){dx} \\ $$$${so}\:\mathrm{2}\int_{−\mathrm{1}} ^{\mathrm{1}} \Upsilon\left({x}\right){dx}=\mathrm{0}\:\:\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \Upsilon\left({x}\right){dx}=\mathrm{0} \\ $$$${hence}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2015}} }{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{\mathrm{2015}}} +\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2015}}} }{dx}\:\:\:\:=\mathrm{0} \\ $$

Commented by Tawa1 last updated on 24/Aug/18

God bless you sir.  what is this sign (Υ) means ??

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{what}\:\mathrm{is}\:\mathrm{this}\:\mathrm{sign}\:\left(\Upsilon\right)\:\mathrm{means}\:?? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Aug/18

its a symbol...but who red marked me and why

$${its}\:{a}\:{symbol}...{but}\:{who}\:{red}\:{marked}\:{me}\:{and}\:{why} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa1 last updated on 24/Aug/18

Owk. Just thinking if the symbol means something.  Maybe the person that redmark you want to like your solution  but mistakenly redmark.    God bless you sir.

$$\mathrm{Owk}.\:\mathrm{Just}\:\mathrm{thinking}\:\mathrm{if}\:\mathrm{the}\:\mathrm{symbol}\:\mathrm{means}\:\mathrm{something}. \\ $$$$\mathrm{Maybe}\:\mathrm{the}\:\mathrm{person}\:\mathrm{that}\:\mathrm{redmark}\:\mathrm{you}\:\mathrm{want}\:\mathrm{to}\:\mathrm{like}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{but}\:\mathrm{mistakenly}\:\mathrm{redmark}. \\ $$$$\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com