Prove that $\int_a^b f(t)f'(t)dt=\frac{1}{2}[(f(b))^2 - (f(a))^2]$ Prove that If $f(t)=u(t)+iv(t)$ is function that derivative continuous on $[a,b]$ then $\displaystyle\int_a^b f(t)f'(t)dt=\frac{1}{2}[(f(b))^2 - (f(a))^2]$
I know If $\displaystyle F'(t)dt=U'(t)dt+iV'(t)dt , \text{then}$
$$\int_a^bF'(t)dt=\int_a^bU'(t)dt+i\int_a^bV'(t)dt =U(b)-U(a)+i(V(b)-V(a))=F(b)-F(a) $$
I'll try to proof the same way but I can't do it. 

Edited
Since $f(t)f'(t)=\frac{1}{2}(f(t)^2)'$
I have $\displaystyle \int_a^b \frac{1}{2}(f(t)^2)'dt=\frac{1}{2}\int_a^b (f(t)^2)'dt=\frac{1}{2}[\int_a^b (u(t)^2)'dt+i\int_a^b (v(t)^2)'dt]=\frac{1}{2}[(u(b)^2)-(u(a)^2)+i\big((v(b)^2)-(v(a)^2)\big)]=\frac{1}{2}[(u(b)^2)+i(v(b)^2)-\big((u(a)^2)+i(v(a)^2)\big)]=\frac{1}{2}[(f(b))^2 - (f(a))^2]$

Please help me to prove this proof , Thank you.  
 A: You can write $$f(t)f'(t) = {1\over 2}(f(t)^2)'$$
Now use Newton-Leibniz formula.
A: $$\int_a^bf(t)f'(t)dt=\int_{f(a)}^{f(b)}udu=\left[\frac u2\right]_{f(a)}^{f(b)}$$
using $u=f(t)\Rightarrow dt=\frac{du}{f'(t)}$
A: Have you ever heard of integration by substitution? It's miraculous how a substitution of a function in x with another value sacrifices a derivative very easily.
So I'll go for the indefinite integral first.
$\displaystyle \int f(t) \cdot f'(t)dx$
$u=f(t)$
$du=f'(t)dt$
$= \displaystyle \int udu$
$=\dfrac{1}{2}u^2+c$
$=\dfrac{1}{2}f^2(x)+c$
$\displaystyle F(b)-F(a)=\displaystyle [\dfrac{1}{2}f^2(x)]_a^b$
$=\displaystyle (\dfrac{1}{2}f^2(b)-\dfrac{1}{2}f^2(a))$
Unless that wasn't the point. But anyway, that's how I'd do it.
http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleIndefinite.aspx
A: Why not simply apply the Fundamental Theorem of Calculus?
If you set $F(x):=\frac{1}{2}(f(x))^2$ then your statement is exactly the same as the Fundamental Theorem of Calculus. (Note that all necessary assumptions to apply the theorem are satisfied).
A: $$\displaystyle\int_a^b f(t)f'(t)dt=\frac{1}{2}[(f(b))^2 - (f(a))^2]$$
$$
\begin{align}
ff'=&(u+iv)(u'+iv')=uu'-vv'+i(uv'+vu')  \\
ff'=&uu'-vv'+i(uv)' \\
\end{align}
$$
$$
\begin{align}
\int ff'dx&= \int uu'-vv'+i(uv)'dx \\
\int ff'dx&= \frac {u^2}2-\frac {v^2}2+iuv \\
\int ff'dx&= \frac 12(u+iv)^2 \\
\int ff'dx&= \frac 12f^2 \\
\end{align}
$$
it gives the same result...
