Answer

Verified

457.8k+ 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.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Why is the Earth called a unique planet class 6 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

One cusec is equal to how many liters class 8 maths CBSE