Antiderivative

If two cars drive along the same road with identical speed at every instant, they cannot drift apart or come together — the distance between them stays fixed forever. Here $F$ and $G$ are the two cars' positions and $f$ is their shared speed; the theorem says the "gap" $G-F$ must be a constant. Crucially this needs $K$ to be a single interval: on two separate intervals, the gap could jump to a different constant on each piece, since you never have to physically travel from one piece to the other.

Define $H(x) = G(x) - F(x)$ for $x\in K$. Since $F$ and $G$ are both antiderivatives of $f$, $H'(x) = G'(x)-F'(x) = f(x)-f(x)=0$ for every $x\in K$: $H$ is differentiable on $K$ with derivative identically zero.

Take any two points $x_1 < x_2$ in $K$ (possible since $K$ is an interval, so the whole segment $[x_1,x_2]$ lies in $K$). $H$ is differentiable, hence continuous, on $[x_1,x_2]$, so the Mean Value Theorem (Lagrange) applies: there is $\xi\in(x_1,x_2)$ with $H(x_2)-H(x_1) = H'(\xi)(x_2-x_1) = 0$, using $H'(\xi)=0$.

Hence $H(x_2)=H(x_1)$ for every pair $x_1<x_2$ in $K$: $H$ takes the same value everywhere on $K$, say $H(x)=C$ for all $x\in K$. Substituting back, $G(x)-F(x)=C$, i.e. $G(x) = F(x) + C$, which is exactly the claim.

Substitution is nothing but the chain rule read backward: differentiating a composite function $F(g(x))$ produces exactly the pattern "outer derivative times inner derivative," $f(g(x))g'(x)$. So whenever an integrand already has this exact shape, recognizing $u=g(x)$ turns a hard-looking integral in $x$ into an easy one in $u$.

Define $\Phi(x) = F(g(x))$. By the chain rule, since $F'=f$, $\Phi'(x) = F'(g(x))\cdot g'(x) = f(g(x))g'(x)$.

This says exactly that $\Phi$ is differentiable with derivative $f(g(x))g'(x)$ at every $x$ in the domain — i.e. $\Phi$ is, by definition, an antiderivative of $f(g(x))g'(x)$.

By the uniqueness theorem above, every antiderivative of $f(g(x))g'(x)$ differs from $\Phi$ by a constant, so $\int f(g(x))g'(x)\,dx = \Phi(x) + C = F(g(x)) + C$, as claimed. In practice one writes $u=g(x)$, $du=g'(x)\,dx$, reducing the left side to $\int f(u)\,du = F(u)+C$ before substituting $u=g(x)$ back at the end.

Integration by parts reverses the product rule: it does not remove the integral, but it shifts the derivative from one factor ($v$) onto the other ($u$), which is exactly the right move when $u$ becomes simpler after differentiating (like $u=x$ becoming $u'=1$) while $v'$ stays easy to antidifferentiate — turning a hard product like $\int x e^x\,dx$ into an easy one.

By the product rule, $(uv)' = u'v + uv'$. Rearranging, $uv' = (uv)' - u'v$.

Take the antiderivative (with respect to $x$, on $K$) of both sides. On the right, $(uv)'$ obviously has antiderivative $uv$ itself (up to a constant, by the uniqueness theorem), so $\int uv'\,dx = \int (uv)'\,dx - \int u'v\,dx = uv - \int u'v\,dx$.

Writing $dv = v'(x)\,dx$ and $du=u'(x)\,dx$ turns this into the compact mnemonic form $\int u\,dv = uv - \int v\,du$. In practice, one factor is chosen as $u$ (to be differentiated) and the rest as $dv$ (to be antidifferentiated), typically preferring to differentiate the factor that simplifies (polynomial, logarithm) and antidifferentiate the factor that stays manageable ($e^x$, $\sin x$, $\cos x$).