Is there any other proof of this?
Second fundamental Theorem of Calculus: If $f$ is differentiable on $[a,b]$ and $f'$ is integrable on $[a,b]$, then $$\int^{b}_a f'(t)dt=f(b)-f(a)$$
My proof
Since $f$ is differentiable on [a,b], then $f'(t)$ exists for all $t\in[a,b]$. For each $n\in \Bbb{N}$, let $P_n$ be an arbitrary partition such that
$$a=x_0<x_1<\cdots<x_n=b.$$
Since $f$ is differentiable on $[a,b]$, then $f$ is continuous on $(a,b).$ So, by Mean Value Theorem, there exists $t_i\in [x_{i-1},x_i]$, for $1\leq i\leq n$ such that
$$ f(x_i)-f(x_{i-1})=f'(t_i)(x_{i-1}-x_i).$$
Summing these up, we have
$$ f(b)-f(a)=\sum^{n}_{i=1}\left[f(x_i)-f(x_{i-1})\right]=\sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i).$$
Integrability of $f'$ on $[a,b]$ implies that
$$ \sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)\leq \sum^{n}_{i=1}f'(t_i)(x_{i-1}-x_i)\leq \sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i),$$
which implies that $$ L(f',P_n)=\sum^{n}_{i=1}m^{f'}_i(x_{i-1}-x_i)\leq f(b)-f(a)\leq \sum^{n}_{i=1}M^{f'}_i(x_{i-1}-x_i)=U(f',P_n),$$
As $n\to\infty,$ $$\lim\limits_{n\to\infty} L(f',P_n)=\lim\limits_{n\to\infty} U(f',P_n)=\int^{b}_a f'(t)dt.$$
Therefore,
$$\int^{b}_a f'(t)dt=f(b)-f(a)$$