# Find the value of the $\rm \int_{-1 }^{1} \left | x \right |dx$

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Find the value of the $\rm \int_{-1 }^{1} \left | x \right |dx$
1. 2
2. 0
3. 3
4. 1

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Correct Answer - Option 4 : 1

Concept:

• $\rm \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$

Calculation:

Given, $\rm \int_{-1 }^{1} \left | x \right |dx$

As we know that, $f\left( x \right) = \left| x \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;\;x < 0}\\ {x,\;\;x \ge 0} \end{array}} \right.$

So, f(x) = - x  when -1 < x < 0 and f(x) = x when 0 < x < 1.

As we know that, $\rm \int_{a}^{b}f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^{b}f(x)dx$

$\rm \Rightarrow\int_{-1 }^{1} \left | x \right |dx=\int_{-1 }^{0} (-x)dx+\int_{0 }^{1} (x)dx$

$\rm \Rightarrow\int_{-1 }^{0} (-x)dx+\int_{0 }^{1} (x)dx=\left [ \frac{-x^{2}}{2} \right ]_{-1}^{0}+\left [ \frac{x^{2}}{2} \right ]_{0}^{1}=\frac{1}{2}+\frac{1}{2}=1$

Hence, the correct option is 4.