Question

Evaluate the given integral.

\(\int\limits_{ - 1}^2 {\left( {x - 2\left| x \right|} \right)dx} \)

Short answer

The value of the given integral is \( - 3.5\)

Step-by-step solution

Definition of function

The definition of absolute value function is as follows:

\(\left| x \right| = \left\{ \begin{aligned}{l}x;x \ge 0\\ - x;x \prec 0\end{aligned} \right\}\)

Rewrite the integral

\(\int\limits_{ - 1}^2 {\left( {x - 2\left| x \right|} \right)dx} = \int\limits_{ - 1}^0 {\left( {x - 2\left| x \right|} \right)dx} + \int\limits_0^2 {\left( {x - 2\left| x \right|} \right)dx} \)

Evaluate the integral

\(\begin{aligned}{l}\int\limits_{ - 1}^2 {\left( {x - 2\left| x \right|} \right)dx &= \int\limits_{ - 1}^0 {\left( {x - 2\left( { - x} \right)} \right)dx} + \int\limits_0^2 {\left( {x - 2\left( x \right)} \right)dx} } \\ &= \int\limits_{ - 1}^0 {\left( {3x} \right)dx} + \int\limits_0^2 {\left( { - x} \right)dx} \\ &= \left. {\frac{{3{x^2}}}{2}} \right)_{ - 1}^0 - \left. {\frac{{{x^2}}}{2}} \right)_0^2\\ &= - \frac{3}{2} - 2\\ &= - 3.5\end{aligned}\)

Therefore, the value of the integral is \( - 3.5\)