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Question Number 26781 by shubhabrata04@gmail.com last updated on 29/Dec/17

Commented by prakash jain last updated on 29/Dec/17

$${\int }_0^1(\log (1-x)-\log x)dx$$$$={\int }_0^1\log (1-x)dx-{\int }_0^1\log xdx$$$$inpartiix=1-udx=-du$$$$={\int }_0^1\log (1-x)dx-{\int }_1^0\log (1-u)(-du)$$$$={\int }_0^1\log (1-x)dx+{\int }_1^0\log (1-u)\left(du\right)$$$$={\int }_0^1\log (1-x)dx-{\int }_0^1\log (1-u)\left(du\right)$$$$=0$$

Commented by malwaan last updated on 29/Dec/17

$$\mathrm{to}\mathrm{Mr}\mathrm{Prakash}$$$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{at}\mathrm{the}\mathrm{forth}$$$$\mathrm{line}\mathrm{the}\mathrm{sign}\mathrm{is}(+)$$$$\mathrm{and}\mathrm{the}\mathrm{answer}\mathrm{is}$$$${\int }_0^1\log (1-\mathrm{x})\mathrm{dx}+$$$${\int }_1^0\log (1-\mathrm{u})\mathrm{du}$$$$=2{\int }_0^0\log (1-\mathrm{x})\mathrm{dx}=0$$$$\mathrm{Am}\mathrm{I}\mathrm{right}???$$

Commented by prakash jain last updated on 29/Dec/17

$$\mathrm{Yes}\mathrm{in}\mathrm{the}\mathrm{next}\mathrm{line}\mathrm{I}\mathrm{swapped}$$$$\mathrm{the}\mathrm{limits}.\mathrm{So}\mathrm{we}\mathrm{are}\mathrm{subtracting}$$$$\mathrm{same}\mathrm{integral}.$$$$$$$${\int }_0^{1}f\left(x\right)dx=-{\int }_1^{0}f\left(x\right)dx$$$$also$$$${\int }_0^{1}f\left(x\right)dx+{\int }_1^{0}f\left(x\right)dx={\int }_0^{0}f\left(x\right)dx$$

Commented by malwaan last updated on 31/Dec/17

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

Answered by lee last updated on 29/Dec/17

$$={\int }_0^1\{\log (1-\mathrm{x})-\mathrm{logx}\}\mathrm{dx}$$$$\mathrm{using}\mathrm{that}\int \mathrm{logxdx}=\mathrm{xlogx}-\mathrm{x}+\mathrm{c}$$$$={[-(1-\mathrm{x})\log (1-\mathrm{x})+(1-\mathrm{x})]}_0^1$$$$-{[\mathrm{xlogx}-\mathrm{x}]}_0^1$$$$=-1+1=0$$

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