Question

Find the value of \[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx\].

Answer 1

Hint: Express the function \[\left| {2x - 1} \right|\] as \[2x - 1\] and \[ - (2x - 1)\] and find the interval and use this to evaluate the integral using the limits.

Complete step-by-step answer:
A modulus function is a function that gives the absolute value of a number or variable.
A definite integral can be divided into a sum of two definite integrals by taking appropriate limits. This property is represented as follows:
\[\int\limits_a^b {f(x)dx} = \int\limits_a^c {f(x)dx} + \int\limits_c^b {f(x)dx} \]
It is defined as follows:
\[f(x) = \left| x \right| = \left\{ \begin{gathered}
  x{\text{ }},x \geqslant 0 \\
   - x,x < 0 \\
\end{gathered} \right..........(1)\]
We have the function \[\left| {2x - 1} \right|\], we need to define this as in equation (1).
We now find the interval where 2x – 1 is less than 0, we have:
\[2x - 1 < 0\]
\[2x < 1\]
\[x < \dfrac{1}{2}\]
Hence, the function \[\left| {2x - 1} \right|\], can be represented as follows:
\[\left| {2x - 1} \right| = \left\{ \begin{gathered}
  2x - 1{\text{ }},x \geqslant \dfrac{1}{2} \\
   - (2x - 1),x < \dfrac{1}{2} \\
\end{gathered} \right............(2)\]
Using definition (2) in the integral and splitting the limits, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = - \int_{ - 1}^{\dfrac{1}{2}} {(2x - 1} )dx + \int_{\dfrac{1}{2}}^2 {(2x - 1} )dx\]
Integral of 1 is x and the integral of x is \[\dfrac{{{x^2}}}{2}\], then, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = - \left\lceil {{x^2} - x} \right\rceil _{ - 1}^{\dfrac{1}{2}} + \left\lceil {{x^2} - x} \right\rceil _{\dfrac{1}{2}}^2\]
Evaluating the limits, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = - \left[ {{{\left( {\dfrac{1}{2}} \right)}^2} - \dfrac{1}{2} - \left( {{{( - 1)}^2} - ( - 1)} \right)} \right] + \left[ {{2^2} - 2 - \left( {{{\left( {\dfrac{1}{2}} \right)}^2} - \dfrac{1}{2}} \right)} \right]\]
Simplifying, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = - \left[ {\dfrac{1}{4} - \dfrac{1}{2} - 2} \right] + \left[ {2 - \left( {\dfrac{1}{4} - \dfrac{1}{2}} \right)} \right]\]
Simplifying further, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = - \left[ { - \dfrac{1}{4} - 2} \right] + \left[ {2 + \dfrac{1}{4}} \right]\]
Taking the negative sign inside the brackets, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = \dfrac{1}{4} + 2 + 2 + \dfrac{1}{4}\]
Simplifying, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = 4 + \dfrac{1}{2}\]
Taking common denominator, we have:
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = \dfrac{{8 + 1}}{2}\]
\[\int_{ - 1}^2 {\left| {2x - 1} \right|} dx = \dfrac{9}{2}\]
Hence, the value of the given integral is \[\dfrac{9}{2}\].

Note: You can also draw the function \[\left| {2x - 1} \right|\] graphically and find the area enclosed by the curve and the x-axis to find the value of the integral.