A calculus problem by Rahil Sehgal

Calculus Level 4

$\large f(x)= \frac{\displaystyle\int_{0}^{k} \lfloor x \rfloor dx}{\displaystyle \int_{0}^{k} \{x\} dx}$

For $f(x)$ as defined above, find $\displaystyle\sum_{k=1}^{n} f(x)$ .

Notations :

$\lfloor \cdot \rfloor$ denotes the floor function .
$\{\cdot \}$ denotes the fractional part function .

n^2 -1

\dfrac{ n(n+1)}{2}

n^2 +1

\dfrac{n^2-n}{2}

1 solution

Aditya Narayan Sharma
Apr 29, 2017

It suffices to evaluate $\displaystyle \int_0^n \lfloor{x\rfloor} \;dx$ to get an answer.

$\displaystyle \begin{aligned} \int_0^n \lfloor{x\rfloor} \;dx & = \sum_{r=0}^{n-1} \int_{r} ^{r+1} r \;dx \\ & =\sum_{r=0}^{n-1} r \\ & = \dfrac{n(n-1)}{2} \end{aligned}$

Exploiting the property $\{x\} =x-\lfloor{x\rfloor}$ it can be easily derived that the ratio is $n-1$

So it follows that $\displaystyle \sum_{n=1}^{n} (n-1)=\dfrac{n^2-n}{2}$

Same way!!!

A Former Brilliant Member - 3 years, 8 months ago

A calculus problem by Rahil Sehgal

1 solution

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