Question

Evaluate: $\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} $.

Answer 1

Hint: First, break the limits from 2 to 3 and 3 to 4, so that the absolute value can be eliminated. After that, add the like terms and integrate them with respect to x. Then apply a limit on it. Now, add the values obtained to get the final result.

Complete step-by-step answer:
Given: - $\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} $ ….. (1)
Break the limits in two parts, so that the absolute value can be eliminated.
For limit 2 to 3, $\left| {x - 2} \right|$ will always return the positive value but $\left| {x - 3} \right|$ and $\left| {x - 4} \right|$ will give negative values.
So, the integration part will be,
$\int_2^3 {\left[ {\left( {x - 2} \right) - \left( {x - 3} \right) - \left( {x - 4} \right)} \right]dx} $ …. (2)
Similarly, for limit 3 to 4, $\left| {x - 2} \right|$ and $\left| {x - 3} \right|$ will always return the positive value but $\left| {x - 4} \right|$ will give a negative value.
So, the integration part will be,
$\int_3^4 {\left[ {\left( {x - 2} \right) + \left( {x - 3} \right) - \left( {x - 4} \right)} \right]dx} $ ….. (3)
Equate the sum of equation (2) and (3) with equation (1),
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = \int_2^3 {\left[ {\left( {x - 2} \right) - \left( {x - 3} \right) - \left( {x - 4} \right)} \right]dx} + \int_3^4 {\left[ {\left( {x - 2} \right) + \left( {x - 3} \right) - \left( {x - 4} \right)} \right]dx} $
Now open the brackets and add or subtract the like terms on the right side of the equation,
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = \int_2^3 {\left( { - x + 5} \right)dx} + \int_3^4 {\left( {x - 1} \right)dx} $
Now integrate the terms,
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = \left[ { - \dfrac{{{x^2}}}{2} + 5x} \right]_2^3 + \left[ {\dfrac{{{x^2}}}{2} - x} \right]_3^4$
Now apply limits on the variables,
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = \left[ {\left( { - \dfrac{{{3^2}}}{2} + 5 \times 3} \right) - \left( { - \dfrac{{{2^2}}}{2} + 5 \times 2} \right)} \right] + \left[ {\left( {\dfrac{{{4^2}}}{2} - 4} \right) - \left( {\dfrac{{{3^2}}}{2} - 3} \right)} \right]$
Now square the terms and multiply the terms and open brackets,
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = \left[ { - \dfrac{9}{2} + 15 + \dfrac{4}{2} - 10} \right] + \left[ {\dfrac{{16}}{2} - 4 - \dfrac{9}{2} + 3} \right]$
Cancel out the common factors and add or subtract the terms,
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = - \dfrac{{18}}{2} + 5 + 2 + 8 - 1$
Cancel out the common factors and add or subtract the terms,
$\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} = 5$

Hence, the value of $\int_2^4 {\left( {\left| {x - 2} \right| + \left| {x - 3} \right| + \left| {x - 4} \right|} \right)dx} $ is 5.

Note: The students might make mistakes by not removing mod by breaking the limits.
Integration can be used to find areas, volumes, central points and many useful things. It is often used to find the area under the graph of a function. The symbol for "Integral" is a stylish "S" (for "Sum", the idea of summing slices).