Question

Integrate!

Answer 1

\[\int\limits_{a}^{b} = 2\]

Answer 2

\[\Huge \int\limits_{0}^{100} \left\{ x \right\}dx\]

Answer 3

its fractional x

Answer 4

\[\{x\}=x-\lfloor x\rfloor=\begin{cases}x&\text{for }0\le x<1\end{cases}\]
and
\(\{x-n\}=\{x\}\) for \(n\in\mathbb{Z}\).

So,
\[\int_0^{100}\{x\}~dx=100\int_0^1x~dx\]

Answer 5

didn't quite get it..

Answer 6

Wait so \[
\{x\} = x-\lfloor x \rfloor
\]?

Answer 7

yes

Answer 8

\[\LARGE 0.21=2.21-[2.21]=2.21-2=0.21\]

Answer 9

Well, \[
\int_0^1\{x\} dx=\int_0^1 x-\lfloor x \rfloor dx =\int_0^1 xdx = x^2\Bigg|_0^1
\]Right?

Answer 10

\[
\int_1^2\{x\} dx=\int_1^2 x-\lfloor x \rfloor dx =\int_1^2 x-1\;dx = \int_0^1udu=u^2\Bigg|_0^1
\]Where \(u=x-1\)

Answer 11

You keep doing this \(100\) times...

Answer 12

\[\int\limits_0^1 x-\lfloor x \rfloor dx =\int\limits_0^1 xdx = x^2\Bigg|_0^1\]
How did this happen?

Answer 13

@DLS So we have
\[\int\limits_0^{100} x-\lfloor x \rfloor dx \]
Let's break the limit
from 0 to 1, we have
\[\int\limits_0^{1} x-\lfloor x \rfloor dx \]
Now tell me, if you have x between 0 to 1
floor of x would be ?

Answer 14

@DLS are you here?

Answer 15

extremely sorry! D: and it would be 0,but still why are we breaking the limit like this?