Answer
(a) Using the Fundamental Theorem of Calculus, we prove that
$G{\rm{''}}\left( t \right) = f\left( t \right)$
(b) By changing the order of integration, we show that
$G\left( t \right) = \mathop \smallint \limits_0^t \left( {t - y} \right)f\left( y \right){\rm{d}}y$
Work Step by Step
We have $G\left( t \right) = \mathop \smallint \limits_0^t \mathop \smallint \limits_0^x f\left( y \right){\rm{d}}y{\rm{d}}x$.
Since $f\left( y \right)$ is a function of $y$ alone, the inner integral is a function of $x$ alone:
$H\left( x \right) = \mathop \smallint \limits_0^x f\left( y \right){\rm{d}}y$
So, $G\left( t \right) = \mathop \smallint \limits_0^t H\left( x \right){\rm{d}}x$.
Using the Fundamental Theorem of Calculus, Part II (Section 5.5) we obtain
$G'\left( t \right) = H\left( t \right)$
Thus,
$G'\left( t \right) = \mathop \smallint \limits_0^t f\left( y \right){\rm{d}}y$
Again using the Fundamental Theorem of Calculus, Part II (Section 5.5) we obtain $G{\rm{''}}\left( t \right) = f\left( t \right)$. Notice that $G\left( t \right)$ is called the "second antiderivative" of $f\left( y \right)$.
(b) Recall the double integral $G\left( t \right) = \mathop \smallint \limits_{x = 0}^t \mathop \smallint \limits_{y = 0}^x f\left( y \right){\rm{d}}y{\rm{d}}x$. We see that the domain is a vertically simple region whose description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le t,0 \le y \le x} \right\}$
Now, we change the order of integration such that the domain becomes a horizontally simple region (please see the figure attached). In this case, the domain description becomes
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le t,y \le x \le t} \right\}$
Thus, by changing the order of the integral we get
$G\left( t \right) = \mathop \smallint \limits_{y = 0}^t \left( {\mathop \smallint \limits_{x = y}^t f\left( y \right){\rm{d}}x} \right){\rm{d}}y$
$ = \mathop \smallint \limits_{y = 0}^t f\left( y \right)\left( {x|_y^t} \right){\rm{d}}y$
$ = \mathop \smallint \limits_0^t \left( {t - y} \right)f\left( y \right){\rm{d}}y$
Hence, $G\left( t \right) = \mathop \smallint \limits_0^t \left( {t - y} \right)f\left( y \right){\rm{d}}y$.
