Question

Sketch the graph of $|x-3|$ and hence evaluate ${\int }_0^6|x-3|dx$

Answer 1

$y=|x-3|=\{\begin{array}{cc}x-3,& \text{if}x\ge 3\\ 3-x,& \text{if}x<3\end{array}$

$\therefore$ Graph of the curve $y=|x-3|$ is

Now, to evaluate ${\int }_0^6|x-3|dx$ , we have to find the area of shaded region as shown below


${\int }_0^6|x-3|dx={\int }_0^3(3-x)dx+{\int }_3^6(x-3)dx$
$={[3x-\frac{x^2}{2}]}_0^3+{[\frac{x^2}{2}-3x]}_3^6$
$=9-\frac{9}{2}+18-18-\frac{9}{2}+9$
$=9$