_ 0 ^ 1 3 ( _ r=0 ^ 101 x + r 3 )dx is equal to (where . represen… - Vidyadip

Question

$\int_{0}^{\frac{1}{3}} (\sum_{r=0}^{101}\{x + \frac{r}{3}\})dx$ is equal to (where {.} represents fractional part function)

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Answer

b
$\int_{0}^{1 / 3}(\{x\}+\left\{x+\frac{1}{3}\right\}+\left\{x+\frac{2}{3}\right\}+\{x+1\}+\left\{x+\frac{4}{3}\right\}+\ldots .$

$\left\{x+\frac{101}{3}\right\}) d x$

$=34 \int_{0}^{1 / 3}\left(\{x\}+\left\{x+\frac{1}{3}\right\}+\left\{x+\frac{2}{3}\right\}\right) d x$

$=34 \int_{0}^{1 / 3}\left(x+x+\frac{1}{3}+x+\frac{2}{3}\right) d x=34 \int_{0}^{1 / 3}(3 x+1) d x$

$=34\left[\frac{3 x^{2}}{2}+x\right]_{0}^{1 / 3}=34\left[\frac{1}{6}+\frac{1}{3}\right]=17$

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