Finding $\int_{-2}^8xf(x)dx$ given $\int_{-2}^8f(x)dx$

I have a continuous function $f:[-2,8]\rightarrow\mathbb{R}$ for which is true that $f(6-x)=f(x)\forall x\in[-2,8]$. Let: $$\int_{-2}^8f(x)dx=10$$ Now, I want to find the: $$\int_{-2}^8xf(x)dx$$ I am thinking of using both the methods of u-substitution and integration by parts, but I need some help. Any ideas?

Here's a cute trick.

If the problem is well-posed, then the solution must be independent of $f$. Therefore, you can take $$ f(x)\equiv1 $$ which is consistent with the hypotheses, and calculate $$ \int_{-2}^8x\ \mathrm dx\equiv 30 $$

Easy peasy!

Using the substitution $w=6-x$, we obtain

\begin{aligned} \int_{-2}^8xf(x)dx&=\int_{-2}^8(6-(6-x))f(6-(6-x))dx\\\\ &=-\int_{8}^{-2}(6-w)f(6-w)dw\\\\ &=\int_{-2}^{8}(6-w)f(6-w)dw\\\\ &=\int_{-2}^{8}(6-w)f(w)dw\\\\ &=6\int_{-2}^{8}f(w)dw-\int_{-2}^{8}wf(w)dw\\\\ &=60-\int_{-2}^{8}xf(x)dx \end{aligned} and thus $$2\int_{-2}^{8}xf(x)dx=60$$ i.e. $$\int_{-2}^{8}xf(x)dx=30.$$