# By using the properties of definite integrals,evaluate the integral$\int\limits_2^8\mid x-5 \mid\;dx$

Toolbox:
• $\int \limits_a^bf(x)dx=F(b)-F(a)$
• $\int \limits_a^cf(x)dx=\int \limits_a^b f(x)dx+\int \limits_b^cf(x)dx$
Given l=$\int\limits_2^8\mid x-5 \mid\;dx$

$|x-5|= \left\{ \begin{array}{1 1} x-5\;&when\;(x-5) \geq 0 & \quad x\geq 5 \\ -(x-5)\;&when\;(x-5) <0 & \quad x<5 \end{array} \right.$

Hence -(x-5)<0 on [2,5] and (x-5)$\geq$0 on [5,8]

Hence $\int\limits_2^5-(x-5)dx+\int\limits_{5}^{8}(x-5)dx$

This is obtained by applying the property

$\int \limits_a^cf(x)dx=\int \limits_a^b f(x)dx+\int \limits_b^cf(x)dx$

On integrating we get,

$\bigg[\frac{x^2}{2}-5x\bigg]_{5}^{2}+\bigg[\frac{x^2}{2}-5x\bigg]_5^8$

On applying limits,

$-\bigg[\bigg(\frac{25}{2}+5 \times 5 \bigg)-\frac{(2)^2}{4}+5 \times 2 ]+\bigg[(\frac{8^2}{2}-5 \times 8)-(\frac{5^2}{2}-5 \times 5)\bigg]$

$=[\frac{25}{2}-25-2+10]+[32-40-\frac{25}{2}+25]$

$=\bigg\{-\bigg[\frac{4}{2}-4-\frac{25}{2}+10\bigg]\bigg\}+\bigg[\frac{25}{2}+10-\frac{4}{2}+4 \bigg]$

$=\frac{9}{2}+\frac{9}{2}=9$

Hence $I=\int\limits_{2}^8\mid x-5 \mid\;dx=9$