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Let $f$ be a real-valued function of a real variable defined as $f(x) = x - [ x ]$  , where  $ [ x ]$  denotes the largest integer less than or equal to $x$.  The value of ${\LARGE \int}_{0.25}^{1.25}  \! f(x) \, \mathrm{d}x$ is _________  (upto $2$ decimal places).
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Answer should be 0.50

a

$\left [ x \right ] =$ Largest  interger  less  than  or  equall  to  x

$\begin{bmatrix} \left [ 0 \leq x < 1 \Rightarrow [x] = 0 \right ] \\ \left [ 1 \leq x < 2 \Rightarrow [x] = 1 \right ] \\ \left [ 2 \leq x < 3 \Rightarrow [x] = 2 \right ]\end{bmatrix}$ & $\begin{bmatrix} \left [ -1 \leq x < 0 \Rightarrow [x] = -1 \right] \\ \left [ -2 \leq x < -1 \Rightarrow [x] = -2 \right ] \\ \left [ -3 \leq x < -2 \Rightarrow [x] = -3 \right ]\end{bmatrix}$

So we can plot  f(x) = [x] as 

 

Now here in $Q^n$,

    f(x) = x- [x]

so I $=\int_{0.25}^{1.25} f(x)dx$

       $=\int_{0.25}^{1.25} (x-[x])dx$

       $=\int_{0.25}^{1.25} xdx$ $-$  $\int_{0.25}^{1.25} [x]dx$

       $=\int_{0.25}^{1.25} xdx$ - $\left [ \int_{0.25}^{1} [x]dx + \int_{1}^{1.25} [x]dx \right ]$      $\begin{Bmatrix}\because  \int_{a}^{c} f(x)dx = \int_{a}^{b} f(x)dx + \int_{b}^{c} f(x)dx  \\ if \ a< b<c\end{Bmatrix}$       

 

since integral of a function gives area under curve so here,

$\int_{0.25}^{1} [x]dx = 0$ {from the above graph}

and $\int_{1}^{1.25} [x]dx = 1 * (1.25 - 1) $

$\because$ in the above graph 

so overall ,

I  $ = \int_{0}^{1.25} xdx   - [0 + 0.25]$

   $=\frac{1}{2}[(1.25)^2 - (0.25)^2] - 0.25$

   $=\frac{1}{2}[(1.25+ 0.25) (1.25 - 0.25)] - 0.25$

   $=\frac{1}{2}*1.50*1 - 0.25$

   $= 0.75 - 0.25$

   $=0.50$

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