Signum+Floor+Integration

Calculus Level 3

$\large \int_{-10}^0 \frac {\left|\frac {2\lfloor x \rfloor}{\lfloor x \rfloor - 3x} \right|}{\frac {2\lfloor x \rfloor}{3x - \lfloor x \rfloor}} dx =\ ?$

Notations:

$\lfloor \cdot \rfloor$ denotes the floor function .
$| \cdot |$ denotes the absolute value function .

The answer is 9.333333333.

1 solution

Chew-Seong Cheong
Mar 17, 2018

Note that the limits of the integral is $x\le 0$ . For $x\le 0$ , $2\lfloor x \rfloor \le 0$ and $\lfloor x \rfloor - 3x \begin{cases} \le 0 \text{ if }x \ge - \frac 13 \\ > 0 \text{ if }x < - \frac 13 \end{cases}$ $\implies \left|\dfrac {2\lfloor x \rfloor}{\lfloor x \rfloor - 3x}\right| = \begin{cases} \dfrac {2\lfloor x \rfloor}{\lfloor x \rfloor - 3x} \text{ if }x \ge - \frac 13 \\ \dfrac {2\lfloor x \rfloor}{3x - \lfloor x \rfloor} \text{ if }x < - \frac 13 \end{cases}$ . Therefore,

$\begin{aligned} I & = \int_{-10}^0 \frac {\left|\frac {2\lfloor x \rfloor}{\lfloor x \rfloor - 3x}\right|}{\frac {2\lfloor x \rfloor}{3x - \lfloor x \rfloor}} dx \\ & = \int_{-\frac 13}^0 \frac {\frac {2\lfloor x \rfloor}{\lfloor x \rfloor - 3x}}{\frac {2\lfloor x \rfloor}{3x - \lfloor x \rfloor}} dx + \int_{-10}^{-\frac 13} \frac {\frac {2\lfloor x \rfloor}{3x - \lfloor x \rfloor}}{\frac {2\lfloor x \rfloor}{3x - \lfloor x \rfloor}} dx \\ & = - \int_{-\frac 13}^0 dx + \int_{-10}^{-\frac 13} dx \\ & = - x\ \bigg|_{-\frac 13}^0 + x\ \bigg|_{-10}^{-\frac 13} \\ & = - 0 - \frac 13 - \frac 13 + 10 \\ & = \frac {28}3 \approx \boxed{9.333} \end{aligned}$

@Ayush Mishra , you can see the LaTex codes by placing your mouse cursor on the formulas. You can also click the pull-down menu " $\cdots$ More" at the bottom of the answer section and select "Toggle LaTex". You should use \ [ and \ ] (no space between the backslash and brackets) instead of \ ( ). Then \ [\int_0^1\ ] becomes $\int_0^1$

If you want to use \ ( \ ) you should add \displaystyle in front, see $\displaystyle \int_0^1$ . Also do not need so many braces { } for example \frac 12 $\frac 12$ , if you want not reduced size fraction \dfrac {11}2 $\dfrac {11}2$ (d for display). or \cfrac \pi 4 $\cfrac \pi 4$ . See I did not use { }. Do not use \left and \right else you will make the brackets too large. See what you codes did Find the value of the integral $\int _{ -10 }^{ 0 }{ \frac { \left| \frac { { 2 }\left\lfloor x \right\rfloor }{ { \left\lfloor x \right\rfloor }-{ { 3 }{ x } } } \right| }{ \left( \frac { { 2 }{ \left\lfloor x \right\rfloor } }{ { 3 }{ x }-{ \left\lfloor x \right\rfloor } } \right) } } { dx }$ . The floor function brackets are too big.

Chew-Seong Cheong - 3 years, 2 months ago

Signum+Floor+Integration

The answer is 9.333333333.

1 solution

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