Question Number 56708 by gunawan last updated on 22/Mar/19
$$\mathrm{for}\:\mathrm{all}\:{n}\:\in\:\mathbb{N},\:{f}_{{n}} \left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\mathrm{for}\:\mathrm{any}\:−\mathrm{1}<{x}<\mathrm{1}.\:\mathrm{If}\:{f}\::\:\left(−\mathrm{1},\mathrm{1}\right)\rightarrow\mathbb{R} \\ $$$$\mathrm{with}\:{f}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{f}_{{n}\:} \mathrm{of}\:\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right)\:{dx}=… \\ $$$$ \\ $$
Commented by Abdo msup. last updated on 22/Mar/19
$${we}\:{have}\:{f}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−{x}\right)^{{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{x}\right)^{{k}} −\mathrm{1} \\ $$$$=\frac{\mathrm{1}−\left(−{x}\right)^{{n}+\mathrm{1}} }{\mathrm{1}+{x}}\:−\mathrm{1}\:=\frac{\mathrm{1}−\left(−{x}\right)^{{n}+\mathrm{1}} −\mathrm{1}−{x}}{\mathrm{1}+{x}} \\ $$$${but}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:\:\:{f}_{{n}} \left({x}\right)=\frac{−{x}}{\mathrm{1}+{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{−{x}}{\mathrm{1}+{x}}\:\Rightarrow\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:=−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:+\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{x}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:+\left[{ln}\mid\mathrm{1}+{x}\mid\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:+{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:. \\ $$$${here}\:{i}\:{have}\:{used}\:{that}\:{lim}\:{f}_{{n}} ={f}\:. \\ $$