I'm reading Gong Sheng's Concise Complex Analysis, where in Chapter 1 reviewing calculus, he says
The fundamental theoren1 of calculus plays the most important role in
calculus. There are two equivalent forms of this theorem:Theorem 1.1 (The Fundo,mental Theorem of Calculus) (differ-form)
Suppose the function $f(x)$ is continuous on $[a, b]$ and
$x\in [a,b]$, Let $$\Phi(x)=\int_a^x f(t)dt$$ , $a\le x \le b$. Then
$\Phi(x)$ is differentiable in $[a, b]$ and $\Phi'(x) = f(x)$,
$d\Phi(x) = f(x)dx$. In other words, if the integral of $f(x)$ is
$\Phi(x)$, then the differential of $\Phi(x)$ is $f(x)$.Theorem 1.2 (The Fundamental Theorem of Calculus) (integral form)
Suppose $\Phi(x)$ is differentiable in $[a, b]$; and $d\Phi(x)/dx$ is
equal to a continuous function $f(x)$. Then
$$\int_a^x f(t) dt = \Phi(x) – \Phi(a)$$
, $(a\le x \le b)$ holds. In other words: If the
differential of $\Phi(x)$ is $f(x)$, then the integral of $f(x)$ is
$\Phi(x)$.By this theorem, finding the integral is the inverse operation of
finding the differential of a function, and properties of
differentials correspond to properties of integrals. For instance:$$\frac{d(f(x)\pm g(x))}{dx}=\frac{df(x)}{dx}\pm\frac{d g(x)}{dx}$$
corresponds to
$$\int (f(x)\pm g(x))dx = \int f(x) dx \pm \int g(x) dx$$ ; and
$$\frac{d}{dx}(fg) = f \frac{dg}{dx}+\frac{df}{dx} g$$
corresponds to $$\int fg' dx = fg – \int gf' dx$$
So far so good… then he mentions that given $y=g(x)$
$$\frac{d f(g(x))}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx}$$
corresponds to $$\int f(g(x))g'(x) dx = \int f(y) dy$$
I couldn't figure out, by the Fundamental Theorem of Calculus, and the differential form $\frac{d f(g(x))}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx}$, how could one derive the integral form $\int f(g(x))g'(x) dx = \int f(y) dy$, or vice versa?
Best Answer
Suppose that $F$ is an anti-derivative of $f$. Then the function $F \circ g$ will have the anti-derivative $(F' \circ g) \cdot g' = (f \circ g) \cdot g'$ by the chain rule. That is, the function $h(x) = F(g(x))$ will have derivative $h'(x) = f(g(x)) g'(x)$.
Thus, if we can find an anti-derivative of $f$, we can use this anti-derivative to find an anti-derivative of $f(g(x)) g'(x)$. This is what $u$-substitution is all about.