Question

Sketch the graph of y=|x+3|and evaluate \(\int_{-6}^{0} |x+3| \,dx\)

Answer 1

The given equation is y=|x+3|

The corresponding values of x and y are given in the following table.

x	-6	-5	-4	-3	-2	-1	0
y	3	2	1	0	1	2	3

On plotting these points, we obtain the graph of y=|x+3| as follows.

It is known that, (x+3)≤0 for -6≤x≤-3 and(x+3)≥0 for -3≤x≤0

∴\(\int_{-6}^{0} |x+3| \,dx\)=-\(\int_{-6}^{-3} |x+3| \,dx\)+\(\int_{-3}^{0} (x+3) \,dx\)

=-\(\bigg[\)\(\frac{x^2}{2}\)+3x\(\bigg]^{-3}_{-6}\)+\(\bigg[\)\(\frac{x^2}{2}\)+3x\(\bigg]^0_{-3}\)

=-\(\bigg[ \bigg(\)\(\frac{(-3)^2}{2}\)+3(-3)\(\bigg)\)-\(\bigg(\)\(\frac{(-6)^2}{2}\)+3(-6)\(\bigg)\bigg]\)+\(\bigg[\)0-(\(\frac{(-3)^2}{2}\)+3(-3)\(\bigg)\bigg]\)

=-\(\bigg[\)\(-\frac{9}{2}\)\(\bigg]\)-\(\bigg[\)\(-\frac{9}{2}\)\(\bigg]\)

=9