I'm trying to show that if $g$ is such that $f(b) - f(a) = \int_a^b g(t) dt$ for any $a<b \in \mathbb{R}$ then we have: (for $f, g \in L^2(\mathbb{R})$)
$$\int_a^b f(t)g(t) = \frac{1}{2}(f(b)^2 - f(a)^2)$$
I can see this would follow if we could treat $g$ as the derivative of $f$ and use integration by parts but I'm not quite sure how to justify this.
Thanks for any help
$$\int_a^b f(x) g(x) dx = \int_a^b \left [ f(a) + \int_a^x g(y) dy\right ] g(x)dx= f(a) \int_a^b g(x)dx + \int_a^b\int_a^x g(y)g(x) dy dx$$ by Fubini's theorem $$\int_a^b\int_a^x g(y)g(x) dy dx = \int_a^b g(y)\int_y^b g(x) dx dy =\int_a^b g(y)(f(b) - f(y)) dy$$ hence $$\int_a^b f(x) g(x) dx = f(a) \int_a^b g(x)dx + f(b) \int_a^b g(y)dy - \int_a^b f(y) g(y) dy$$ and we can replace $y$ by $x$ because it is a dummy variable $$2 \int_a^b f(x) g(x) dx = f(a) (f(b) - f(a)) + f(b) (f(b) - f(a))$$ $$\int_a^b f(x) g(x) dx = \frac{1}{2} (f(b)^2 - f(a)^2)$$
