# How do I prove that this integral from a to b is equivalent to the second statement?

## If f' is continuous on [a, b], show that 2int_a^bf(x)f'(x)dx)= [f(b)]^2-[f(a)]^2

$2 {\int}_{a}^{b} \setminus f \left(x\right) \setminus f ' \left(x\right) \setminus \setminus \mathrm{dx}$
$= 2 {\int}_{a}^{b} \setminus \frac{d}{\mathrm{dx}} {\left(\frac{1}{2} f \left(x\right)\right)}^{2} \setminus \setminus \mathrm{dx}$
$= {\int}_{a}^{b} \setminus \frac{d}{\mathrm{dx}} {\left(f \left(x\right)\right)}^{2} \setminus \setminus \mathrm{dx}$
$= {\left[{\left(f \left(x\right)\right)}^{2}\right]}_{a}^{b}$
$= {\left(f \left(b\right)\right)}^{2} - {\left(f \left(a\right)\right)}^{2}$