How can you find $\int_0^8xf'(x)\,{\rm d}x$ when $f(8) = 9$ and $f(4+x) = -f(4-x)$?

Question

This is the actual question:

If $f: \mathbb R \to \mathbb R$ is a continuous differientable function and given that $f(8) = 9$ and $f(4+x) = -f(4-x)$, what is $\int_0^8xf'(x)$d$x$?

This is what I have thus far:

If $f(8) = 9$ and $f(4+x) = -f(4-x)$, then $-f(0) = 9 \Rightarrow f(0) = -9$.

Say $u = x$ and d$v = f'(x)$d$x$, then d$u = $d$x$ and $v=f(x)$.

So $\int_0^8xf'(x)$d$x = \int_0^8u$d$v = uv|^8_0 - \int^8_0v$d$u. = (8\cdot9 - 0\cdot9) - F(x)|^8_0 = 72 - F(x)|^8_0$ (not sure if this is a correct notation). Since I don't know what $f$ is, I obviously don't know its integral either. What other way can I use to find the solution?

Thanks in advance.

Answer 1

HINT.-$$\int_0^8 xf'(x){\rm d}x =\int_0^8 xd(f(x)= [xf(x)]_0^8 - \int_0^8 f(x){\rm d}x\\\int_0^8 xf'(x){\rm d}x=72-\int_0^8 f(x){\rm d}x$$ Now calculation gives $f(4)=0$ and because of $f(4+x) = -f(4-x)$ the function can be considered as being odd respect to the new origin $x=4$.

A realization of such a function is the straight line $$f(x)=9(\frac x4-1)$$ In this case we have $$\int_0^8 xf'(x){\rm d}x=72-\int_0^89(\frac x4-1)dx=72$$

It seems it could be other realizations of $f(x)$ according to the fact that $f$ odd respect of $x=4$ but always the answer will be $72$.

Answer 2

The function $f$ is odd with respect to the point $x=4$. We therefore replace $f$ with $$g(x):=f(4+x)\ .$$ Then $$g(-x)= -g(x),\quad g(4)=9,\quad f(x)=g(x-4),\quad f'(x)=g'(x-4)\ .$$ It follows that $$J:=\int_0^8 x f'(x)\>dx=\int_0^8 x g'(x-4)\>dx\ ,$$ and substituting $x:=4+t$ $(-4\leq t\leq 4)$ gives $$J=\int_{-4}^4 (4+t) g'(t)\>dt=4\bigl(g(4)-g(-4)\bigr)+\int_{-4}^4 tg'(t)\>dt=72+0=72\ ,$$ since $g$ is odd and $g'$ is even.

Answer 3

Solution

Notice that $$\int_0^8xf'(x)\,{\rm d}x=\int_0^8x\,{\rm d}f(x)=xf(x)\bigg|_0^8-\int_0^8 f(x)\,{\rm d}x,$$

where$$xf(x)\bigg|_0^8=8f(8)-0f(0)=72,$$

and \begin{align*} \int_0^8 f(x)\,{\rm d}x&=\int_0^4 f(x)\,{\rm d}x+\int_4^8 f(x)\,{\rm d}x\\ &=\int_0^4 f(x)\,{\rm d}x+\int_{-4}^0 f(x+8)\,{\rm d}x\\ &=\int_0^4 f(x)\,{\rm d}x-\int_{-4}^0 f(-x)\,{\rm d}x\\ &=\int_0^4 f(x)\,{\rm d}x+\int_{4}^0 f(x)\,{\rm d}x\\ &=0. \end{align*} where we apply the fact $f(x+8)=-f(-x)$ and a series of variable substitution.

As a result,$$\int_0^8xf'(x)\,{\rm d}x=72-0=72.$$