# Define a function \[f:R \to R{\text{ }}by{\text{ }}f(x) = \max \left\{ {\left| x \right|,\left| {x - 1} \right|,.......\left| {x - 2n} \right|} \right\}\]

Where n is a fixed natural number. Then \[\int\limits_0^{2n} {f(x)dx} \] is.

\[(A){\text{ }}n{\text{ }}\]

\[(B){\text{ }}{n^2}\]

\[(C){\text{ }}3n\]

\[(D){\text{ }}3{n^2}\]

Answer

Verified

366.6k+ views

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.

Last updated date: 03rd Oct 2023

â€¢

Total views: 366.6k

â€¢

Views today: 9.66k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is meant by shramdaan AVoluntary contribution class 11 social science CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Is current density a scalar or a vector quantity class 12 physics JEE_Main

An alternating current can be produced by A a transformer class 12 physics CBSE

What is the value of 01+23+45+67++1617+1819+20 class 11 maths CBSE