# Prove: using the Fundamental theorem of calculus

I have trouble doing proofs using the fundamental theorem of calculus and I think seeing an example would help.

Suppose we have a continuous function $f$ defined on a real interval and a function $F$ that is also defined on the same interval; and $F(x)=\int^b_x fdt$. Then for $x$ in that interval, $F'(x)=-f(x)$.

I know this can be done using the fundamental theorem of calculus; how is it applied?

By the way, the theorem says this:

Let $f$ be integrable on $[a,b]$. For $x\in[a,b]$ put $F(x)=\int^x_a f(t)dt$. Then $F$ is continuous on $[a,b]$ and furthermore if $f$ is continuous at a point $x_0$ of $[a,b]$ then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.

We name the function from FTC $F$ and our new function $G$ then $$G(x) = \int_x^b f(t)\ \mathrm dt = \underbrace{\int_a^b f(t)\ \mathrm dt}_{\text{constant}} - \int_a^x f(t) \ \mathrm dt = C - F(x)$$ So $G'(x) = -F'(x) = -f(x)$ by FTC.
• @Rubi Because it's an integral independent of $x$. It has constant bounds ($a$ and $b$) and a function independent of $x$. Mar 4, 2015 at 19:34