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Hint:- Check values of \[f(x)\] in different ranges from \[\left[ {0,2n} \right]\].

As we know that we had to find the value of \[\int\limits_0^{2n} {f(x)dx} \].

\[ \Rightarrow \]Where, \[f(x) = \max \left\{ {\left| x \right|,\left| {x - 1} \right|,.......\left| {x - 2n} \right|} \right\}{\text{ }}\] (1)

So, from equation 1, we can see that the value of \[f(x)\] will be,

\[ \Rightarrow f(x) = \left| {x - 2n} \right|\] for x in range \[\left[ {0,n} \right]\] (2)

\[ \Rightarrow f(x) = \left| x \right|\] for x in range \[\left[ {n,2n} \right]\] (3)

So, to find the value of \[\int\limits_0^{2n} {f(x)dx} \] we had to break the limits of the integral into two parts,

\[ \Rightarrow \]So, \[\int\limits_0^{2n} {f(x)dx} = \int\limits_0^n {f(x)dx} + \int\limits_n^{2n} {f(x)dx} {\text{ }}\] (4)

\[ \Rightarrow \]And, \[\int\limits_0^n {f(x)dx} = \int\limits_0^n {\left| {x - 2n} \right|dx} \] from equation 2

As, we know that \[2n{\text{ }} > x\] for x in range \[\left[ {0,n} \right]\].

\[ \Rightarrow \]So, \[\int\limits_0^n {f(x)dx} = \int\limits_0^n {\left( {2n - x} \right)dx} \]

Now we had to solve the above equation.

So, solving the above equation. It becomes,

\[ \Rightarrow \int\limits_0^n {f(x)dx} = \left( {2nx - \dfrac{{{x^2}}}{2}} \right)_0^n = \left( {2{n^2} - \dfrac{{{n^2}}}{2}} \right) = \dfrac{{3{n^2}}}{2}\] (5)

\[ \Rightarrow \]And, \[\int\limits_n^{2n} {f(x)dx} = \int\limits_n^{2n} {\left| x \right|dx} \] from equation 3

\[ \Rightarrow \]So, \[\int\limits_n^{2n} {f(x)dx} = \int\limits_n^{2n} {\left( x \right)dx} \] for x in range \[\left[ {n,2n} \right]\]

So, solving the above equation. It becomes,

\[ \Rightarrow \int\limits_n^{2n} {f(x)dx} = \left( {\dfrac{{{x^2}}}{2}} \right)_n^{2n}{\text{ = }}\left( {\dfrac{{4{n^2}}}{2} - \dfrac{{{n^2}}}{2}} \right){\text{ = }}\dfrac{{3{n^2}}}{2}\] (6)

Now, putting the value of \[\int\limits_0^n {f(x)dx} \] and \[\int\limits_n^{2n} {f(x)dx} \] from equation 5 and 6 to equation 4.

\[ \Rightarrow \]So, \[\int\limits_0^{2n} {f(x)dx} = \dfrac{{3{n^2}}}{2} + \dfrac{{3{n^2}}}{2} = 3{n^2}\]

Hence, the correct option will be D.

Note:- Whenever we came up with this type of problem then we should break the

limits of integral according to the value of function in different ranges. Otherwise

solving the given integral without checking the value of integral in different integral

will give us incorrect answers.

As we know that we had to find the value of \[\int\limits_0^{2n} {f(x)dx} \].

\[ \Rightarrow \]Where, \[f(x) = \max \left\{ {\left| x \right|,\left| {x - 1} \right|,.......\left| {x - 2n} \right|} \right\}{\text{ }}\] (1)

So, from equation 1, we can see that the value of \[f(x)\] will be,

\[ \Rightarrow f(x) = \left| {x - 2n} \right|\] for x in range \[\left[ {0,n} \right]\] (2)

\[ \Rightarrow f(x) = \left| x \right|\] for x in range \[\left[ {n,2n} \right]\] (3)

So, to find the value of \[\int\limits_0^{2n} {f(x)dx} \] we had to break the limits of the integral into two parts,

\[ \Rightarrow \]So, \[\int\limits_0^{2n} {f(x)dx} = \int\limits_0^n {f(x)dx} + \int\limits_n^{2n} {f(x)dx} {\text{ }}\] (4)

\[ \Rightarrow \]And, \[\int\limits_0^n {f(x)dx} = \int\limits_0^n {\left| {x - 2n} \right|dx} \] from equation 2

As, we know that \[2n{\text{ }} > x\] for x in range \[\left[ {0,n} \right]\].

\[ \Rightarrow \]So, \[\int\limits_0^n {f(x)dx} = \int\limits_0^n {\left( {2n - x} \right)dx} \]

Now we had to solve the above equation.

So, solving the above equation. It becomes,

\[ \Rightarrow \int\limits_0^n {f(x)dx} = \left( {2nx - \dfrac{{{x^2}}}{2}} \right)_0^n = \left( {2{n^2} - \dfrac{{{n^2}}}{2}} \right) = \dfrac{{3{n^2}}}{2}\] (5)

\[ \Rightarrow \]And, \[\int\limits_n^{2n} {f(x)dx} = \int\limits_n^{2n} {\left| x \right|dx} \] from equation 3

\[ \Rightarrow \]So, \[\int\limits_n^{2n} {f(x)dx} = \int\limits_n^{2n} {\left( x \right)dx} \] for x in range \[\left[ {n,2n} \right]\]

So, solving the above equation. It becomes,

\[ \Rightarrow \int\limits_n^{2n} {f(x)dx} = \left( {\dfrac{{{x^2}}}{2}} \right)_n^{2n}{\text{ = }}\left( {\dfrac{{4{n^2}}}{2} - \dfrac{{{n^2}}}{2}} \right){\text{ = }}\dfrac{{3{n^2}}}{2}\] (6)

Now, putting the value of \[\int\limits_0^n {f(x)dx} \] and \[\int\limits_n^{2n} {f(x)dx} \] from equation 5 and 6 to equation 4.

\[ \Rightarrow \]So, \[\int\limits_0^{2n} {f(x)dx} = \dfrac{{3{n^2}}}{2} + \dfrac{{3{n^2}}}{2} = 3{n^2}\]

Hence, the correct option will be D.

Note:- Whenever we came up with this type of problem then we should break the

limits of integral according to the value of function in different ranges. Otherwise

solving the given integral without checking the value of integral in different integral

will give us incorrect answers.

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