Question

Evaluate \int\limits_2^4 {\left\\{ {\left| {x - \left. 2 \right|} \right. + \left| {\left. {x - 3} \right| + \left| {\left. {x - 4} \right|} \right.} \right.} \right\\}} dx

Answer 1

Here we use the modulus operation, we can define the modulus.
The basic definition for modulus function is $f(x) = \left| {\left. x \right|} \right.$,
Now, if $x$ is non-negative, then the output of $f$ will be $x$ itself.
If $x$ is negative then the output of $f$ will be the magnitude of $x$, that is we can write as it $- x$
In this question we have to split the terms upon their limits.
We have to find one term of splitting the given integral and similarly do this all the integral.
Finally we get the answer.

Formula used: Integration property,$\int\limits_a^a {f(x)} = 0$
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
$\int {dx = x}$

Complete step-by-step answer:
It is given the integration is \int\limits_2^4 {\left\\{ {\left| {x - \left. 2 \right|} \right. + \left| {\left. {x - 3} \right| + \left| {\left. {x - 4} \right|} \right.} \right.} \right\\}} dx
Now we can write to split the integrals $\int\limits_2^4 {\left| {\left. {x - 2} \right|} \right.} dx + \int\limits_2^4 {\left| {\left. {x - 3} \right|} \right.} dx + \int\limits_2^4 {\left| {\left. {x - 4} \right|} \right.} dx....\left( 1 \right)$
First we take $\int\limits_2^4 {\left| {\left. {x - 2} \right|} \right.} dx....\left( 2 \right)$
We can use the modulus definition for $\int\limits_2^4 {\left| {\left. {x - 2} \right|} \right.} dx$
$\Rightarrow f(x - 2) = \left| {\left. {x - 2} \right|} \right.$,
Now, if $x - 2$ is greater than or equal to $2$, that is $x - 2$ nonnegative, then the output of $f$ will be $x - 2$ itself.
If $x - 2$ is less $2$ than the output of $f$ will be the magnitude of $x - 2$, which can be write as $- (x - 2)$
The interval will be greater than or equal to $2$ for that, the interval limit will be $2$ to $4$
The interval will for less than $2$ for that, the interval limit will be $2$ to $2$
Then we have to split the equation $\left( 2 \right)$ limits $\int\limits_2^2 { - (x - 2)} dx + \int\limits_2^4 {(x - 2)dx} ....\left( 3 \right)$
Here we use, integration property for same limit $\int\limits_a^a {f(x)} = 0$
Now we take the first terms, $\int\limits_2^2 { - (x - 2)} dx = 0....\left( 4 \right)$
Now equation $\left( 3 \right)$ becomes $0 + \int\limits_2^4 {(x - 2)dx}$
Also we split it as, $\int\limits_2^4 {xdx - 2} \int\limits_2^4 {dx}$
Here we use the formula for integration,
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$
$\int {dx = x}$
Now we can write it as,$\left( {\dfrac{{{x^2}}}{2}} \right)_2^4 - 2(x)_2^4$
Substitute the integration limit value in $x$, in the form of $\left( {{\text{upper limit - lower limit}}} \right)$
So we can write it as, $\Rightarrow \left( {\dfrac{{{4^2}}}{2} - \dfrac{{{2^2}}}{2}} \right) - 2(4 - 2)$
On squaring the numerator terms and subtract the bracketed terms we get,
$\int\limits_2^4 {(x - 2)dx} = \left( {\dfrac{{16}}{2} - \dfrac{4}{2}} \right) - 2(2)$
Let us divided and multiply it,
$\Rightarrow \left( {8 - 2} \right) - 4$
On subtract the terms,
$\Rightarrow 6 - 4 = 2...\left( 5 \right)$
$\therefore$ Substitute $\left( 4 \right)$ and $\left( 5 \right)$ in $\left( 3 \right)$ becomes, $\int\limits_2^2 { - (x - 2)} dx + \int\limits_2^4 {(x - 2)dx} = 2$
Hence, $\int\limits_2^4 {\left| {\left. {x - 2} \right|} \right.} dx = 2....\left( a \right)$
Now, we take the second term in equation $\left( 1 \right)$,
$\Rightarrow \int\limits_2^4 {\left| {\left. {x - 3} \right|} \right.} dx....\left( 6 \right)$
We can use the modulus definition for $\int\limits_2^4 {\left| {\left. {x - 3} \right|} \right.} dx$
$\Rightarrow f(x - 3) = \left| {\left. {x - 3} \right|} \right.$
Now, if $x - 3$ is greater than or equal to $3$, then the output of $f$ will be $x - 3$ itself.
If $x - 3$ is less $3$ than the output of $f$ will be the magnitude of $x - 3$, which can be write as - $$$$(x - 3)
The interval will for greater than or equal to $3$ for that, the interval limit will be $3$ to $4$
The interval will for less than $3$ for that, the interval limit will be $2$ to $3$
Now we split equation $\left( 6 \right)$ limits
$\int\limits_2^3 { - (x - 3)} dx + \int\limits_3^4 {(x - 3)dx}$
We split the terms by using the limits we get,
$- \int\limits_2^3 {xdx + 3} \int\limits_2^3 {dx} + \int\limits_3^4 {xdx - 3} \int\limits_3^4 {dx}$
Here we use the formula for integration, we get
$- \left( {\dfrac{{{x^2}}}{2}} \right)_2^3 + 3(x)_2^3 + \left( {\dfrac{{{x^2}}}{2}} \right)_3^4 - 3(x)_3^4$
Substitute the integration interval value in $x$ $\left( {{\text{upper limit - lower limit}}} \right)$
$- \left( {\dfrac{9}{2} - \dfrac{4}{2}} \right) + 3(3 - 2) + \left( {\dfrac{{16}}{2} - \dfrac{9}{2}} \right) - 3(4 - 3)$
Here, the base is same, so we take LCM,
$- \left( {\dfrac{{9 - 4}}{2}} \right) + 3(1) + \left( {\dfrac{{16 - 9}}{2}} \right) - 3(1)$
On subtract the numerator terms we get,
$- \dfrac{5}{2} + 3 + \dfrac{7}{2} - 3$
Cancel the same terms ,
- \dfrac{5}{2} + 3 + \dfrac{7}{2} - 3$$$$ = 1
Hence equation $\left( 6 \right)$ becomes,
$\int\limits_2^4 {\left| {\left. {x - 3} \right|} \right.} dx = 1....\left( b \right)$
Now, we take the third terms of the equation $\left( 1 \right)$
$\Rightarrow \int\limits_2^4 {\left| {\left. {x - 4} \right|} \right.} dx...\left( 7 \right)$
We can use the modulus definition for $\int\limits_2^4 {\left| {\left. {x - 4} \right|} \right.} dx$
$\Rightarrow f(x - 4) = \left| {\left. {x - 4} \right|} \right.$
Now, if $x - 4$ is greater than or equal to $4$, then the output of $f$ will be $x - 4$ itself. If $x - 4$ is less $4$ then the output of $f$ will be the magnitude of, which can be write as $-$ $(x - 4)$
The interval will for greater than or equal to $4$ for that, the interval limit will be 4 to $4$
The interval will for less than 4 for that, the interval limit will be $2$ to 4
Now we can split the equation $\left( 7 \right)$ by their limits we get,
$\int\limits_2^4 { - (x - 4)} dx + \int\limits_4^4 {(x - 4)dx} ....\left( 8 \right)$
By integration property $\int\limits_a^a {f(x)} = 0$
Then, \int\limits_4^4 {(x - 4)dx} $$$$ = 0
Now we take the first terms of the equation \left( 8 \right) $$$\int\limits_2^4 { - (x - 4)} dx$$ $$ + 0$$ We split the terms by using the limits $$ - \int\limits_2^4 {xdx + } 4\int\limits_2^4 {dx} $$ Here we use the formula for integration, we get $$ - \left( {\dfrac{{{x^2}}}{2}} \right)_2^4 + 4(x)_2^4$$ Substitute the integration interval value in $$x$$ $$\left( {{\text{upper limit - lower limit}}} \right)$$ $$ - \left( {\dfrac{{16}}{2} - \dfrac{4}{2}} \right) + 4(4 - 2)$$ Let us divided the numerator terms and subtract the bracket term we get, $$ - (8 - 2) + 4(2)$$ On subtracting, $$ - (6) + 8$$$$ = 2$$ Hence equation \left( 8 \right)becomes, $$\int\limits_2^4 {\left| {\left. {x - 4} \right|} \right.} dx = 2....\left( c \right)$$ \therefore $Equation$\left( 1 \right), becomes $$ \Rightarrow \int\limits_2^4 {\left| {\left. {x - 2} \right|} \right.} dx + \int\limits_2^4 {\left| {\left. {x - 3} \right|} \right.} dx + \int\limits_2^4 {\left| {\left. {x - 4} \right|} \right.} dx$$ Substitute \left( a \right)$$\left( b \right)$and$\left( c \right)$in equation$\left( 1 \right)$ we get,
2 + 1 + 2$$$$ = 5
Hence
\int\limits_2^4 {\left| {\left. {x - 2} \right|} \right.} dx + \int\limits_2^4 {\left| {\left. {x - 3} \right|} \right.} dx + \int\limits_2^4 {\left| {\left. {x - 4} \right|} \right.} dx$$$$ = 5

Note: Here, the derivation of the sum is little big, but the concept is small.
If the integrative involves the limits in interval $x = 0$, then the integration does not exist.
By the definition of modulus function, a modulus function is a function which gives us the magnitude of that number.