If $$\int_{-a}^{a} (|x| + |x - 2|) dx = 22$$, $$a > 2$$ and $$[x]$$ denotes the greatest integer $$\leq x$$, then $$\int_{-a}^{a} (x + |x|) dx$$ is equal to ______

Given,

$$\int_{-a}^{a}(|x|+|x-2|)\,dx=22, \qquad a>2$$

We first simplify the integrand.

Since

$$a>2,$$

split according to the critical points

$$x=0$$ and $$x=2.$$

For

$$-a\le x<0,$$

$$|x|=-x, \qquad |x-2|=2-x$$

Hence,

$$|x|+|x-2|=2-2x$$

For

$$0\le x<2,$$

$$|x|=x, \qquad|x-2|=2-x$$

Hence,

$$|x|+|x-2|=2$$

For

$$2\le x\le a,$$

$$|x|=x, \qquad |x-2|=x-2$$

Hence,

$$|x|+|x-2|=2x-2$$

Therefore,

$$\int_{-a}^{a}(|x|+|x-2|)\,dx$$

$$= \int_{-a}^{0}(2-2x)\,dx +\int_{0}^{2}2\,dx +\int_{2}^{a}(2x-2)\,dx$$

Now,

$$\int_{-a}^{0}(2-2x)\,dx =[2x-x^2]_{-a}^{0}$$

$$=2a+a^2$$

Also,

$$\int_0^2 2\,dx=4$$ and$$\int_2^a(2x-2)\,dx=[x^2-2x]_2^a$$

$$=a^2-2a$$

Hence,

$$2a+a^2+4+a^2-2a=22$$

$$2a^2+4=22$$

$$2a^2=18$$

$$a^2=9$$

Since

$$a>2,$$

$$a=3$$

Now evaluate

$$\int_{-a}^{a}(x+|x|)\,dx$$

For

$$x<0,$$

$$x+|x|=x-x=0$$

For

$$x\ge0,$$

$$x+|x|=x+x=2x$$

Thus,

$$\int_{-3}^{3}(x+|x|)\,dx=\int_{-3}^{0}0\,dx+\int_0^3 2x\,dx$$

$$=[x^2]_0^3$$

$$=9$$

Hence, the required value is

$$\boxed{9}$$