Let $$f(x) = \max\{|x+1|, |x+2|, \ldots, |x+5|\}$$. Then $$\int_{-6}^{0} f(x)dx$$ is equal to ______

We need $$\int_{-6}^{0} f(x)\,dx$$ where $$f(x) = \max\{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$$.

The functions $$|x+k|$$ for $$k = 1,2,3,4,5$$ have V-shaped graphs with vertices at $$x=-k$$, and each measures the distance from $$x$$ to $$-k$$. Since we seek the maximum distance from $$x$$ to one of the points $$\{-1,-2,-3,-4,-5\}$$, this maximum will occur at whichever endpoint, $$-1$$ or $$-5$$, is farther away from $$x$$.

The midpoint of $$-1$$ and $$-5$$ is $$-3$$. For $$x\le -3$$ the farthest point is $$-1$$, so $$f(x)=|x+1|=-(x+1)=-x-1$$. For $$x\ge -3$$ the farthest point is $$-5$$, so $$f(x)=|x+5|=x+5$$.

To verify at the transition point $$x=-3$$, we compute $$|-3+1|=2,\quad|-3+2|=1,\quad|-3+3|=0,\quad|-3+4|=1,\quad|-3+5|=2$$, and observe that both $$|x+1|$$ and $$|x+5|$$ equal 2, confirming that $$x=-3$$ is indeed the transition point.

Next we compute the integral over $$[-6,-3]$$:

$$\int_{-6}^{-3}(-x-1)\,dx=\left[-\frac{x^2}{2}-x\right]_{-6}^{-3}=\left(-\frac{9}{2}+3\right)-\left(-\frac{36}{2}+6\right)=\left(-\frac{3}{2}\right)-(-12)=-\frac{3}{2}+12=\frac{21}{2}.$$

Then we compute the integral over $$[-3,0]$$:

$$\int_{-3}^{0}(x+5)\,dx=\left[\frac{x^2}{2}+5x\right]_{-3}^{0}=(0)-\left(\frac{9}{2}-15\right)=0-\left(-\frac{21}{2}\right)=\frac{21}{2}.$$

Adding these two parts gives

$$\int_{-6}^{0}f(x)\,dx=\frac{21}{2}+\frac{21}{2}=21.$$

The answer is $$\boxed{21}$$.