why $\int_{0}^ {x} = \int_{\frac{-1}{2}}^{y}$ and $\int_{0}^{x} = \int_{y}^{1} ?$ I have some confusion in integration . My confusion marked in red and green circle as given below
Im not getting  why $$\int_{0}^ {x} = \int_{\frac{-1}{2}}^{y}$$ and $$\int_{0}^{x} = \int_{y}^{1} $$ ?
Im not getting  how its derive ?
 A: Hint: Try to plot these areas and look firstly from $Ox$, then from $Oy$ axes.
More formally: we should prove
$$\left\{ \begin{array}{} 
0 \leqslant x \leqslant 1 \\
0 \leqslant y \leqslant x 
\end{array} \right\} = \left\{ \begin{array}{}
0 \leqslant y \leqslant 1 \\
y \leqslant x \leqslant 1
\end{array} \right\}
$$
and then obtain
$$\int\limits_{0}^{1}\int\limits_{0}^{x} dxdy= \int\limits_{0}^{1}\int\limits_{y}^{1}dydx$$
Same for second.
A: In the integral in your problem, the bounds are $0<y<x$, so, as @zkutch wrote, if you plot the graph $y=f(x)=x$, the area will be the lower triangle if you split the unit square $([0,1]\times[0,1])$ with this function. The same area corresponds to the bounds $y<x<1$. Do the same with $[-\frac{1}{2},0]$ interval.
To verify that the integrals can be interchanged, Fubini-Tonelli's theorem must be verified:
$$
\int_{B}f = \int_{B}f^{+} + \int_{B}f^{-}
$$
Since your function is bounded on the compact set, both of these integrals are finite, so integrals can be interchanged.
