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:

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.