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?
 A: You're given:
$$I=\int_{-2}^8xf(x)dx$$
Use the $a+b-x$ property on this definite integral to get:
$$\begin{align}
I&=\int_{-2}^8 (6-x)\cdot f(6-x)dx
\\
&=\int_{-2}^8 (6-x)\cdot f(x)dx \tag{$\because f(6-x)=f(x)$ given}
\\
&=6\int_{-2}^8f(x)-I
\end{align}$$ 
and you can solve it from here.
A: 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!
A: 
Note the following formula we always have,
  $$\color{red}{\int_a^bg(x)dx= \int_a^bg(a+b-x)dx}$$

Then with $a=-2,~b=8$ and given that $f(x) = f(6-x) $ we get $$I= \int_{-2}^8 xf(x)dx= \int_{-2}^8 (6-x)f(6-x)dx=6\int_{-2}^8 f(x)dx-\int_{-2}^8 xf(x)dx\\=60-I$$
hence solving for I we obtain,  $$I=\int_{-2}^8 xf(x)dx=30$$
A: 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.
$$\int_{-2}^{8}xf(x)dx=30.$$
A: $$I:=\int_{-2}^8 x f(x)dx=-\int_8^{-2} (6-x) f(6-x)dx=\int_{-2}^8 (6-x) f(6-x)dx$$ so that
$$I+I=\int_{-2}^8 (x+6-x)f(6-x)dx=6\int_{-2}^8f(x)dx.$$
