Answer
(a) it makes sense
(b) it does not make sense
(c) it does not make sense
(d) it makes sense
Work Step by Step
(a) $\mathop \smallint \limits_0^1 \mathop \smallint \limits_1^x f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Writing as an iterated integral, we have $\mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 1}^x f\left( {x,y} \right){\rm{d}}y} \right){\rm{d}}x$.
We hold $x$ constant and evaluate the inner integral with respect to $y$. This gives us a function of $x$ alone:
$S\left( x \right) = \mathop \smallint \limits_{y = 1}^x f\left( {x,y} \right){\rm{d}}y$
Integrating the resulting function $S\left( x \right)$ with respect to $x$ we obtain the value of the double integral. So, it makes sense.
(b) $\mathop \smallint \limits_0^1 \mathop \smallint \limits_1^y f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Writing as an iterated integral, we have $\mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 1}^y f\left( {x,y} \right){\rm{d}}y} \right){\rm{d}}x$.
The inner integral does not make sense since we integrate with respect to $y$, but the upper limit is $y$.
(c) $\mathop \smallint \limits_0^1 \mathop \smallint \limits_x^y f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Writing as an iterated integral, we have $\mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = x}^y f\left( {x,y} \right){\rm{d}}y} \right){\rm{d}}x$.
The inner integral does not make sense since we integrate with respect to $y$, but the upper limit is $y$.
(d) $\mathop \smallint \limits_0^1 \mathop \smallint \limits_x^1 f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Writing as an iterated integral, we have $\mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = x}^1 f\left( {x,y} \right){\rm{d}}y} \right){\rm{d}}x$.
We hold $x$ constant and evaluate the inner integral with respect to $y$. This gives us a function of $x$ alone:
$S\left( x \right) = \mathop \smallint \limits_{y = x}^1 f\left( {x,y} \right){\rm{d}}y$
Integrating the resulting function $S\left( x \right)$ with respect to $x$ we obtain the value of the double integral. So, it makes sense.
