Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 114304 by mnjuly1970 last updated on 18/Sep/20

$$calculus.....$$$$evaluate:$$$$\mathrm{\Omega }={\int }_{-1}^{1}xln(1^x+2^x+3^x+6^x)dx=???$$$$$$

Commented by MJS_new last updated on 18/Sep/20

$$=\underset{-1}{\stackrel{1}{\int }}x\ln \left((1+2^x)(1+3^x)\right)dx=$$$$=\underset{-1}{\stackrel{1}{\int }}x\ln (1+2^x)dx+\underset{-1}{\stackrel{1}{\int }}x\ln (1+3^x)dx$$$$\mathrm{looks}\mathrm{easier}\mathrm{now}...$$

Commented by Dwaipayan Shikari last updated on 18/Sep/20

$$\frac{log\left(6\right)}{3}$$

Commented by Dwaipayan Shikari last updated on 18/Sep/20

$${\int }_{-1}^{1}xlog(1+2^x)+xlog(1+3^x)dx$$$$=I_a+I_b$$$$I_a={\int }_{-1}^{1}xlog(1+2^x)={\int }_{-1}^1-xlog(1+2^x)+xlog{2}^x=I_a$$$$2I_a=2{\int }_0^1{x}^{2}log\left(2\right)$$$$I_a=\frac{log\left(2\right)}{3}$$$$I_b=\frac{log\left(3\right)}{3}$$$$I_a+I_b=\frac{log\left(6\right)}{3}$$

Commented by MJS_new last updated on 18/Sep/20

$$\mathrm{yesss}!\mathrm{I}\mathrm{had}\mathrm{been}\mathrm{focused}\mathrm{on}\mathrm{the}\mathrm{integral},$$$$\mathrm{forgot}\mathrm{the}\mathrm{symmetric}\mathrm{borders}...$$

Commented by mnjuly1970 last updated on 18/Sep/20

$$grateful..=$$

Commented by mnjuly1970 last updated on 18/Sep/20

$$thankyou....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com