Integral of a piecewise function 
Check whether the integral $\int_{-1}^2{f(x)[2x+1]}dx$ exists where 
  $$f(x)=
\begin{cases}
x+2, &\text{$x<0$}\\
1, & \text{$x=0$}\\
xe^{x^2},& \text{$x>0$}
\end{cases}$$
  and $[x]$ denotes the greatest integer function. If it exists then calculate the integral.

I'm clueless about how to do the whole proof by definition stuff. Is there a way to prove that this function is integrable without any epsilon/delta and partition proofs?
 A: Because of the fact that the discontinuities of $f(x)$ are of measure zero, the function is Riemann integrable. A proof of this exists here. This means we can validly take the integral of $f(x)$ on either side of the discontinuity and sum those:
$$\int_{-1}^{2} f(x)[2x+1]dx = \int_{-1}^{0}(x+2)[2x+1]dx + \int_{0}^{2}xe^{x^2}[2x+1]dx$$
Next, because $[2x+1]$ is the same value on for any x in the interval $[n, n+\frac{1}{2})$ where $n \in \mathbb{Z}$ and these new discontinuities are also of measure zero, we can seperate this into a larger sum with each value of $[2x+1]$ written out.
$$-1 \cdot \int_{-1}^{-\frac{1}{2}} (x+2)dx + 0 \cdot \int_{-\frac{1}{2}}^{0} (x+2)dx + 1 \cdot \int_{0}^{\frac{1}{2}} xe^{x^2}dx + 2 \cdot \int_{\frac{1}{2}}^{1} xe^{x^2}dx + 3 \cdot \int_{1}^{\frac{3}{2}} xe^{x^2}dx + 4 \cdot \int_{\frac{3}{2}}^{2} xe^{x^2}dx$$
Now we remove the term that is multiplied by $0$ and find $\int_{a}^{b} xe^{x^2}dx$ by setting $u=x^2$ and finding that $du=2xdx$.
$$-\int_{-1}^{-\frac{1}{2}} (x+2)dx + \frac{1}{2}\int_{0}^{\frac{1}{4}} e^{u}du + \int_{\frac{1}{4}}^{1} e^{u}du + \frac{3}{2}\int_{1}^{\frac{9}{4}} e^{u}du + 2\int_{\frac{9}{4}}^{4} e^{u}du$$
Finally we take the antiderivatives and evaluate.
$$-\left[\frac{x^2}{2}+2x\right]_{-1}^{-\frac{1}{2}}+\frac{1}{2}[e^u]_{0}^{1/4}+[e^u]_{1/4}^{1}+\frac{3}{2}[e^u]_{1}^{9/4}+2[e^u]_{9/4}^4$$
$$=-\left(\frac{1}{8}-1-\frac{1}{2}+2\right)+\frac{e^{1/4}-1}{2}+\left(e-e^{1/4}\right)+\frac{3e^{9/4}-3e}{2}+\left(2e^4-2e^{9/4}\right)$$
$$=-\frac{9}{8}-\frac{e^{1/4}}{2}-\frac{e}{2}-\frac{e^{9/4}}{2}+2e^4$$
$$=\frac{16e^4-4e^{9/4}-4e-4e^{1/4}-9}{8} \approx 101.326$$
