States the inverse relation between differentiation and integration. (just like addition and subtraction) and any continuous function has an antiderivative
If f is continuous on [a, b] and F(X) is an antiderivative of f, then
$$ \int_a^bf(x)dx=F(x)|_a^b=F(b)-F(a) $$
it yields the Net change.
If $f$ is continuous on an open interval I containing a [therefore a is a constant], then, for every x in the interval, (t is a variable, x is not used to avoid confusion)
$\frac d {dx}[\int_a^xf(t)dt]=f(x)$ and $\frac d {dx}[\int_a^uf(t)dt]=f(u)u’$
This pretty intuitive,
$\frac d {dx}[\int_a^xf(t)dt]=\frac d {dx}[F(x)-F(a)]$ by the FTOC1
$\frac d {dx}[F(x)-F(a)]=\frac d {dx} F(x)$ because of F(a) is constant
$\frac d {dx} F(x)=f(x)$ by the definition of antiderivative.
$\frac d {dx}[\int_a^uf(t)dt]=\frac d {dx} [F(u)-F(a)]$ by the FTOC1
$\frac d {dx}[F(u)-F(a)]=\frac d {dx} F(u)$ because of F(a) is constant
$\frac d {dx} F(u)=f(u)u'$ by the definition of antiderivative and chain rule
Or if you want a more intuitive proof, check it out on the heading above