# By using the properties of definite integrals,evaluate the integral$\int\limits_{-5}^5\mid x+2 \mid\;dx$

Toolbox:
• $\int \limits_a^bf(x)dx=F(b)-F(a)$
• $f(x)=|x+a|=f(x)= \left\{ \begin{array}{1 1} x+a\;&when\;x+a \geq 0 & \quad x\geq -a \\ -(x+a)\;&when\;x+a <0 & \quad x<-0 \end{array} \right.$
• $\int \limits_a^cf(x)dx=\int \limits_a^b f(x)dx+\int \limits_b^cf(x)dx$
Given $I=\int\limits_{-5}^5\mid x+2 \mid\;dx$
$|x+2|=f(x)= \left\{ \begin{array}{1 1} x+2\;&when\;(x+2) \geq 0 & \quad x\geq -2 \\ -(x+2)\;&when\;(x+2) <0 & \quad x<-2 \end{array}\right.$
Hence the limitsare -5 to -2 and -2 to 5
Hence $\int\limits_{-5}^5\mid x+2 \mid\;dx=\int\limits_{-5}^{-2}\mid x+2 \mid\;dx+\int\limits_{-2}^5\mid x+2 \mid\;dx$
$=\int\limits_{-5}^{-2}-(x+2)\;dx+\int\limits_{-2}^{5}(x+2)\;dx$
On integrating we get,
$\bigg[\frac{x^2}{2}+2x\bigg]_{-5}^{-2}+\bigg[\frac{x^2}{2}+2x\bigg]_{-2}^5$
On applying limits,
$-\bigg[\bigg(\frac{(-2)^2}{2}+2(-2)\bigg)-\bigg(\frac{(-5)^2}{2}+2 \times (-5)\bigg)\bigg]$
$+\bigg[\bigg(\frac{5^2}{2}+2 \times 5 \bigg)-\bigg(\frac{(-2)^2}{2}-2 \times (-2)\bigg)\bigg]$
$=\bigg\{-\bigg[\frac{4}{2}-4-\frac{25}{2}+10\bigg]\bigg\}+\bigg[\frac{25}{2}+10-\frac{4}{2}+4 \bigg]$
$=-2+4+\frac{25}{2}-10+\frac{25}{2}+10-2+4$
Hence $I=\int\limits_{-5}^5\mid x+2 \mid\;dx=29$
edited Jan 31, 2014 by yamini.v