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