Question

A function $f$ is defined by $f\left(x\right)=\frac{1}{2^{r-1}},\frac{1}{2r}<x\le \frac{1}{2^{r-1}},r=1,2,3,.....$ then the value of ${\int }_0^{1}f\left(x\right)dx$ is equal

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{2}{3}$

${\int }_0^{1}f\left(x\right)dx=\sum _{r=1}^{\infty }{\int }_{2^{-r}}^{2^{-(r-1)}}\frac{1}{2^{r-1}}dx$
$=\sum _{r=1}^{\infty }\frac{1}{2^{r-1}}[2^{-(r-1)}-2^{-r}]$
$=\sum _1^{\infty }{2}^{-2(r-1)}-\sum _1^{\infty }{2}^{-2r+1}$
$=(2^2-2)\sum _1^{\infty }{2}^{-2r}=2\cdot \frac{1}{4}\cdot \frac{1}{1-1/4}=\frac{2}{3}\cdot$