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Question: A function ƒ is defined by ƒ(x) = \(\frac{1}{2^{r - 1}}\), \(\frac{1}{2^{r}}\)\< x £\(\frac{1}{2^{r ...

A function ƒ is defined by ƒ(x) = 12r−1\frac{1}{2^{r - 1}}, 12r\frac{1}{2^{r}}< x £12r−1\frac{1}{2^{r - 1}}, r = 1, 2, 3…. then the value of ∫01ƒ(x)\int_{0}^{1}{ƒ(x)} dx is equal –

A

1/3

B

¼

C

2/3

D

1/3

Answer

2/3

Explanation

Solution

∫01ƒ(x)\int_{0}^{1}{ƒ(x)}dx = 12r−1\frac{1}{2^{r - 1}}dx = ∑r=1∞12r−1\sum_{r = 1}^{\infty}\frac{1}{2^{r - 1}} [2 – (r – 1) – 2–r]

= ∑1∞2−2(r−1)\sum _ { 1 } ^ { \infty } 2 ^ { - 2 ( r - 1 ) } – ∑1∞2−2r+1\sum_{1}^{\infty}2^{- 2r + 1}

= (22 – 2) ∑1∞2−2r\sum _ { 1 } ^ { \infty } 2 ^ { - 2 r } = 2 . 14\frac { 1 } { 4 } . 11−1/4\frac { 1 } { 1 - 1 / 4 } = 23\frac { 2 } { 3 }