DLS
  • DLS
Integrate!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Compassionate
  • Compassionate
\[\int\limits_{a}^{b} = 2\]
DLS
  • DLS
\[\Huge \int\limits_{0}^{100} \left\{ x \right\}dx\]
DLS
  • DLS
its fractional x

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anonymous
  • anonymous
\[\{x\}=x-\lfloor x\rfloor=\begin{cases}x&\text{for }0\le x<1\end{cases}\] and \(\{x-n\}=\{x\}\) for \(n\in\mathbb{Z}\). So, \[\int_0^{100}\{x\}~dx=100\int_0^1x~dx\]
DLS
  • DLS
didn't quite get it..
anonymous
  • anonymous
Wait so \[ \{x\} = x-\lfloor x \rfloor \]?
DLS
  • DLS
yes
DLS
  • DLS
\[\LARGE 0.21=2.21-[2.21]=2.21-2=0.21\]
anonymous
  • anonymous
Well, \[ \int_0^1\{x\} dx=\int_0^1 x-\lfloor x \rfloor dx =\int_0^1 xdx = x^2\Bigg|_0^1 \]Right?
anonymous
  • anonymous
\[ \int_1^2\{x\} dx=\int_1^2 x-\lfloor x \rfloor dx =\int_1^2 x-1\;dx = \int_0^1udu=u^2\Bigg|_0^1 \]Where \(u=x-1\)
anonymous
  • anonymous
You keep doing this \(100\) times...
DLS
  • DLS
\[\int\limits_0^1 x-\lfloor x \rfloor dx =\int\limits_0^1 xdx = x^2\Bigg|_0^1\] How did this happen?
ash2326
  • ash2326
@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 ?
ash2326
  • ash2326
@DLS are you here?
DLS
  • DLS
extremely sorry! D: and it would be 0,but still why are we breaking the limit like this?
anonymous
  • anonymous
|dw:1378658537164:dw|
DLS
  • DLS
oh! well explained,thanks man <3

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