1 ∫ (2^x / 2^(3x)) dx 0
[m] =∫ ^{1}_{0}2^{x-3x}dx=∫ ^{1}_{0}2^{-2x}dx=-\frac{1}{2}∫ ^{1}_{0}2^{-2x}d(-2x)=-\frac{1}{2}\frac{2^{-2x}}{ln2}|^{1}_{0}=-\frac{1}{2ln2}\cdot (2^{-2}-2^{0})=-\frac{1}{2ln2}\cdot (\frac{1}{4}-1)=\frac{3}{8ln2}[/m]
