If f(x)=12nwhen 12n+1

∫1/2n1 f(x)dx=∫1/2n1/2n−1 f(x)dx+∫1/2n−11/2n−2 f(x)dx+⋯+∫1/21 f(x)dx=12n−112n−1−12n+12n−212n−2−12n−1+⋯+11−12=122n−1+122n−3+⋯+12=121−1/22n1−1/22=231−14nlimn→∞ ∫1/2n1 f(x)dx=23