Fundamental Theorem of Calculus and discontinuous functions Will the Fundamental Theorem of Calculus be useful in evaluating: 
$$y=\int_{0.5}^{1} \left \lfloor t \right \rfloor dt?$$
The Riemann sums make this evaluation easy. I, however, would like to know how( if possible), can I use the theorem to evaluate this definite integral. Here, I assumed that $y$ is continuous in $[0.5,1]$ and differentiable in $(0,1)$. 
Edit: To be clear about what I wish to know, I ask if I could evaluate the definite integral just by taking the difference, $y(1)-y(0.5)$( second part of the Fundamental Theorem of Calculus), where $y$ is the antiderivative of $\left \lfloor  x\right \rfloor$. 
I'm not sure if I'm allowed to talk about antiderivatives of discontinuous functions. Am I allowed to? 
 A: I would suggest you to have a look at this answer where I have discussed the Fundamental Theorem of Calculus for functions which are not necessarily continuous. 
Next note that the integrand here has a discontinuity at $x=1$ and as far as the interval of integration is concerned the discontinuity is removable and hence we just remove it by changing value of integrand at $x=1$ to $0$. Doing this has no impact on the value of the integral (because the value of a Riemann integral does not depend on the value of the integrand at a finite number of points in the interval under consideration) and thus the desired integral is equal to $\int_{1/2}^{1}0\,dx=0$.

In general if we have to evaluate an integral of the form $\int_{a} ^{b} f(x) \, dx$ where $f$ has a finite number of discontinuities at $x_{1},x_{2},\dots,x_{n}$ in $[a, b] $ with $$a\leq x_{1}\leq x_{2}\leq \dots\leq x_{n} \leq b$$ such that left and right hand limits of $f$ exist at each point of discontinuity then we split the integral as follows $$\int_{a} ^{b} f(x) \, dx=\int_{a} ^{x_{1}}f(x)\,dx+\int_{x_{1}}^{x_{2}}f(x)\,dx+\cdots+\int_{x_{n-1}}^{x_{n}}f(x)\,dx+\int_{x_{n}}^{b}f(x)\,dx$$ Each integral on right can be evaluated by observing that the integrand has removable discontinuity at the end points $x_{i}$ and the integrand can be redefined at these points to make it continuous on that interval.
You can use this technique to evaluate $\int_{1}^{2}[x^{2}]\,dx$ as $$\int_{1}^{2}[x^{2}]\,dx=\int_{1}^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{\sqrt{3}}2\,dx+\int_{\sqrt{3}}^{2}3\,dx$$

Also to answer your question about evaluating the integral under consideration via second part of fundamental theorem of calculus, note that there is no anti-derivative of $[x] $ on interval $[1/2,1]$ (why? perhaps you should answer this yourself, but let me know if you feel issue here) and hence we can't use fundamental theorem of calculus here.
Also note that continuous functions are guaranteed to have an anti-derivative, but continuity is not necessary. There are discontinuous functions which possess anti-derivative and if we have a function $f$ which is Riemann integrable on $[a, b] $ and possesses an anti-derivative $F$ (note that existence of anti-derivative is not guaranteed and hence we are assuming its existence here as a part of the hypotheses) such that $F'(x) =f(x) $ for all $x\in[a, b] $ then we have $\int_{a}^{b}f(x)\,dx=F(b)-F(a)$.
A: $y $ is continuous at $[0.5,1]  \implies $
$$\int_{0,5}^1\lfloor t\rfloor dt=
\lim_{x\to 1^-}\int_{0,5}^x\lfloor t \rfloor dt =0. $$
A: By definitión $[x]$ is defined as the biggest integer notexceding $x$, for example $[\pi]=3$ and $[-1/2]=-1$. The floor function $x \mapsto [x]$ is continuous in the interval $[1/2,1)$; so $$\int_{1/2}^1 [x]dx=\lim_{t \to 1^-} \int_{1/2}^t [x]dx.$$ Also, for $1/2\leq x<1$ we have that $[x]=0$, so $$\int_{1/2}^1 [x]dx=\lim_{t \to 1^-} \int_{1/2}^t [x]dx=\lim_{t \to 1^-} \int_{1/2}^t 0dx=\lim_{t \to 1^-}0=0$$
A: Yes you can talk about the anti derivatives of discontinuous functions. But if a function $f$ has a jump discontinuity (like the greatest integer function $[x] $ in question which has jump discontinuity at integers) then it does not possess an anti-derivative and hence it's integral can not evaluated by using Leibniz Rule (that is difference between values of anti-derivative at the end points of the interval).
There are discontinuous functions which don't have jump discontinuity and then they may possess anti-derivative. For example the function $f(x)=2x\sin(1/x)-\cos(1/x),f(0)=0$ is continuous everywhere except at $0$. It possesses an anti-derivative $g(x) =x^{2}\sin(1/x),g(0)=0$ and for all real $a, b$ we have $$\int_{a} ^{b} f(x) \, dx=g(b) - g(a) $$ However in cases where the integrand is having jump discontinuity in the interval of integration there is still some hope. Here we can not use anti-derivative (because it does not exist) but we can split the interval of integration into multiple intervals via the points of discontinuity and evaluate integral on each subinterval and add these integrals to get the value of the integral on the given interval. Splitting the interval of integration into multiple sub-intervals has the advantage that the function has discontinuity only at the end points and these are removable. So the function can be modified accordingly at end points to make it continuous and the integral evaluated as usual.
For the current question the integrand already has discontinuity on end point of the interval so no need to split the interval and integral is directly evaluated as $\int_{1/2}^{1}0\,dx=0$.
