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

Question

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?

Answer 1

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. $$\int_{-2}^{8}xf(x)dx=30.$$

Answer 2

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.

Answer 3

$$I:=\int_{-2}^8 x f(x)dx=-\int_8^{-2} (6-x) f(6-x)dx=\int_{-2}^8 (6-x) f(6-x)dx$$ so that

$$I+I=\int_{-2}^8 (x+6-x)f(6-x)dx=6\int_{-2}^8f(x)dx.$$

Answer 4

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, $$I=\int_{-2}^8 xf(x)dx=30$$

Answer 5

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!