What is $\int_a^{b}(f'(t))^2 dt$? Let $f:[a,b] \to \mathbb R$ be  nice differentiable function.  By Fundamental Theorem of Calculus (FTC), we have $$\int_a^{b} f'(t) dt = f(b) -f(a).$$
By Cauchy-Schwartz inequality, we have, $\int_{a}^{b} |f(t)| dt \leq  (b-a)^{1/2}(\int_a^b |f(t)|^2 dt)^{1/2} .$  Using this, and FTC, we have $\int_{a}^{b} (f'(t))^2 dt \geq  (b-a)^{-1} (f(b)-f(a))^2.$ 

Question: (1) Can compute $\int_a^{b}(f'(t))^2 dt$ precisely?  (2) Can we improve inequality $\int_{a}^{b} (f'(t))^2 dt \geq (b-a)^{-1} (f(b)-f(a))^2$ ?

 A: In answer to (1), let $f(x)=e^{x^2/2}$. Then $(f'(x))^2=(xe^{x^2/2})^2=x^2e^{x^2}$. This has no elementary antiderivative, as we have
$$
\int x^2e^{x^2} \, dx=\int x(xe^{x^2} \, dx)= \frac{1}{2}xe^{x^2}-\frac{1}{2}\int e^{x^2}\, dx
$$
and this last integral is well-known not to be elementary.
It follows that there can be no general elementary formula for $\int (f'(x))^2 \, dx$.
A: In reverse order: If
$$
f(x) = f(a) + \underbrace{\frac{f(b) - f(a)}{b - a}}_{m}(x - a),\quad a \leq x \leq b,
$$
then $f'(t) \equiv m$, so
$$
\int_{a}^{b} f'(t)^{2}\, dt = m^{2} (b - a)
= \frac{\bigl(f(b) - f(a)\bigr)^{2}}{b - a}.
$$
In words, the stated inequality is saturated for affine functions (as would be expected from Cauchy-Schwarz), so without further constraints on $f$ there is no tighter lower bound.
As for evaluating the integral of $(f')^{2}$, presumably that means something like "in terms of $a$, $b$, $f(a)$, $f(b)$, $f'(a)$, $f'(b)$". The answer is (tentatively, pending your exact meaning, but pretty firmly) no: Letting $\phi$ be a function that vanishes to second order at $a$ and $b$ (e.g., $(x - a)^{2} (x - b)^{2}$ multiplied by a smooth function), and replacing $f$ by $f + \phi$ does not change the value of any functional of the endpoint values, but does ("usually") change the integral of the derivative squared.
