MCQ
$\int_{0}^{\frac{1}{3}} (\sum_{r=0}^{101}\{x + \frac{r}{3}\})dx$ is equal to (where {.} represents fractional part function)
- A$7$
- ✓$17$
- C$27$
- D$37$
$\left\{x+\frac{101}{3}\right\}) d x$
$=34 \int_{0}^{1 / 3}\left(\{x\}+\left\{x+\frac{1}{3}\right\}+\left\{x+\frac{2}{3}\right\}\right) d x$
$=34 \int_{0}^{1 / 3}\left(x+x+\frac{1}{3}+x+\frac{2}{3}\right) d x=34 \int_{0}^{1 / 3}(3 x+1) d x$
$=34\left[\frac{3 x^{2}}{2}+x\right]_{0}^{1 / 3}=34\left[\frac{1}{6}+\frac{1}{3}\right]=17$
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