Twentieth integral

Calculus Level 3

Given the equation:

$2016= \displaystyle \int_{0}^{x} \left \lfloor t \right \rfloor \, dt$

Solve for $x$ .

The answer is 64.

1 solution

A Former Brilliant Member
Mar 8, 2016

$\displaystyle \int_{0}^{x} \left \lfloor t \right \rfloor dt = 2016$
$\therefore \displaystyle \sum_{n=0}^{x-1}n\int_{n}^{n+1}1 dt = 2016$
$\therefore \displaystyle \sum_{n=0}^{x-1} n = 2016$
Using, $\displaystyle \sum_{r=0}^{n} r = \dfrac{n \cdot(n+1)}{2}$
$\dfrac{(x-1)\cdot(x)}{2} = 2016$
$x\cdot(x-1) = 4032$
$x\cdot(x-1) = 64\cdot63$
$\therefore x = 64$

Can you explain your first step? Thanks.

Nihar Mahajan - 5 years, 3 months ago

$\displaystyle \int_{0}^{x}\lfloor t \rfloor dt = \int_{0}^{1}\lfloor t \rfloor dt + \int_{1}^{2}\lfloor t \rfloor dt + \ldots + \int_{x-1}^{x}\lfloor t \rfloor dt$
When $t \in (n,n+1) , \lfloor t \rfloor = n$

A Former Brilliant Member - 5 years, 3 months ago

Also, the integral is the area under the graph, which consists of rectangles of width = 1 unit, and height = n units, where n goes from [0,x]

A Former Brilliant Member - 5 years, 3 months ago

I don't completely agree with the solution. There is another one: -63. I don't see why x must be positive...

Diana Ferreira - 1 year, 1 month ago

Twentieth integral

The answer is 64.

1 solution

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