Question

Sketch the graph of y = |x + 3| and evaluate ${\displaystyle {\int }_{-6}^0|x+3|dx}$

Answer 1

y = |x + 3|

At x = -3, y = 0

AQ is the line y = x + 3

When x + 3 < 0,

y = -(x + 3) = -x – 3

The graph of the line AP is AP.

∴ y =|x + 3| is shown in the graph.

${\displaystyle {\int }_{-6}^1|x+3|dx}$

${\displaystyle ={\int }_{-6}^{-3}|-x-3|dx+{\int }_{-3}^0(x+3)dx}$

${\displaystyle ={[-\frac{x^2}{2}-3x]}_{-6}^{-3}+{[\frac{x^2}{2}+3x]}_{-3}^0}$

${\displaystyle =-{[\frac{x^2}{2}+3x]}_{-6}^{-3}+ {[\frac{x^2}{2}+3x]}_{-3}^0}$

${\displaystyle =-{[\frac{x^2}{2}+3x]}_{-6}^{-3}+{[\frac{x^2}{2}+3x]}_{-3}^0}$

${\displaystyle =-[(\frac{9}{2}-9)-(\frac{36}{2}-18)+0-(\frac{9}{2}-9)]}$

${\displaystyle =-[-\frac{9}{2}-0]+\frac{9}{2}}$

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

= 9 square unit