If $f(a)=f(b)=0$, Show that $\int_a^b xf(x)f'(x)dx=-\frac 12\int[f(x)]^2dx$

Question

$\int_a^b xf(x)f'(x)dx=\frac {-1}2\int[f(x)]^2dx$

My Work:

1) Rearrangement of terms:$$\int_a^b [f(x)][xf'(x)]dx=\frac {-1}2\int_a^b[f(x)][f(x)]dx$$

2) Mean value Theorem of integration: $$f(z)\int_a^b xf'(x)dx=\frac {-f(z)}2\int_a^b f(x)dx$$

3) Cancellation of terms: $$\int_a^b xf'(x)dx=\frac {-1}2\int_a^b f(x)dx$$

4) LHS Integration by parts:$$u'(x)=f'(x)$$ $$v(x)=x$$ $$(f(x)x)|^b_a-\int_a^b f(x)dx=\frac {-1}2\int_a^b f(x)dx$$

5) Integration: $$f(b)b-f(a)a-[F(b)-F(a)]=\left(\frac {-1}2\right)[F(b)-F(a)]$$

6) Substitution $f(a)=f(b)=0$: $$[F(a)-F(b)]=\left(\frac 12\right)[F(a)-F(b)]$$

The end result does not make any sense, was there an error in my thought process?

Attack the problem directly, integration by parts, $v(x)=x$, $u'(x)=f(x)f'(x)$. You will finish very quickly. — André Nicolas, Dec 06 '15 at 22:23
How do I determine $u(x)$? Is it simply $u(x)=F(x)f(x)$? This doesn't make sense to me because of the product rule — Kevin McDonough, Dec 07 '15 at 06:07

score 0 · Accepted Answer · answered Dec 06 '15 at 22:26

Your mistake is in the "Mean Value Theorem of Integration". Even if there is a theorem that says what you want to use (I'm not aware), there is no reason for the $z$ on each side to be the same.

As for solving the problem, as André has mentioned, it is a straightforward integration by parts.

If $f(a)=f(b)=0$, Show that $\int_a^b xf(x)f'(x)dx=-\frac 12\int[f(x)]^2dx$

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