Question

$\int\limits^2_0[x] \, dx$ where [.] represents greatest integer function of x

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int_0^2\rm [x] \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int_0^1\rm [x] \: dx + \displaystyle\int_1^2\rm [x] \: dx$

We know,

$\boxed{ \rm{ \:[x] = 0 \: \: \: \: if \: x \: \in \: [0,1) \: }}$

and

$\boxed{ \rm{ \:[x] = 1 \: \: \: \: if \: x \: \in \: [1,2) \: }}$

So, using this, above integral can be rewritten as

$\rm \: = \: \displaystyle\int_0^1\rm 0 \: dx + \displaystyle\int_1^2\rm 1 \: dx$

$\rm \: = \: 0 + \bigg[x\bigg]_1^2$

$\rm \: = \: 2 - 1$

$\rm \: = \: 1$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\int_0^2\rm [x] \: dx \: = \: 1 \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{ \rm{ \:\displaystyle\int_a^b\rm f(x)dx \: = \: \displaystyle\int_a^b\rm f(y)dy \: }}$

$\boxed{ \rm{ \:\displaystyle\int_a^b\rm f(x)dx \: = \: - \: \displaystyle\int_b^a\rm f(x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_a^b\rm f(x)dx = \displaystyle\int_a^b\rm f(a + b - x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_0^a\rm f(x)dx \: = \: \displaystyle\int_0^a\rm f(a - x)dx \: }}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^a\rm f(x)dx \: = \:2 \displaystyle\int_0^a\rm f(x)dx \: \: if \: f( - x) = f(x) }}$

$\boxed{ \rm{ \:\displaystyle\int_{ - a}^a\rm f(x)dx \: = \:0 \: \: if \: f( - x) = - \: f(x) }}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{2a}\rm f(x)dx \: = \:0 \: \: if \: f(2a - x) = - \: f(x) }}$

$\boxed{ \rm{ \:\displaystyle\int_{0}^{2a}\rm f(x)dx \: = \:\displaystyle\int_0^a\rm f(x)dx \: \: if \: f(2a - x) = f(x) }}$

Answer 2

Step-by-step explanation:

Let,

$\displaystyle\rm I=\int \limits_{0}^{2}\left[x\right] d x .$

Now

$\displaystyle \text{Let }\rm I=\int_{0}^{2}[x] d x=\int_{0}^{1}[x] d x+\int_{1}^{2}[x] d x$

$\\ \displaystyle \rm=\int_{0}^{1} 0 d x+\int_{1}^{2} 1 d x$

$\\ \rm \bigg[\because[x]= \rm \left\{\begin{array}{l}0,0 \leq x<1 \\ 1,1 \leq x<2\end{array}\right]$

$\begin{array}{l}\rm =0+(x)_{1}^{2}=2-1=1 \\ \\ \boxed{\color{orangered}\displaystyle \rm\therefore \int_{0}^{2}[x] d x=1} \end{array}$