Answer
$\mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^4 \left( {{x^3} + {y^2} + z} \right){\rm{d}}x{\rm{d}}y{\rm{d}}z = 693$
Work Step by Step
Evaluate
$\mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^4 \left( {{x^3} + {y^2} + z} \right){\rm{d}}x{\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{y = 1}^4 \left( {\left( {\frac{1}{4}{x^4} + x{y^2} + xz} \right)|_2^4} \right){\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{y = 1}^4 \left( {64 + 4{y^2} + 4z - 4 - 2{y^2} - 2z} \right){\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{y = 1}^4 \left( {60 + 2{y^2} + 2z} \right){\rm{d}}y{\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^3 \left( {\left( {60y + \frac{2}{3}{y^3} + 2yz} \right)|_1^4} \right){\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^3 \left( {240 + \frac{{128}}{3} + 8z - 60 - \frac{2}{3} - 2z} \right){\rm{d}}z$
$ = \mathop \smallint \limits_{z = 0}^3 \left( {222 + 6z} \right){\rm{d}}z$
$ = \left( {222z + 3{z^2}} \right)|_0^3 = 666 + 27 = 693$
So, $\mathop \smallint \limits_{z = 0}^3 \mathop \smallint \limits_{y = 1}^4 \mathop \smallint \limits_{x = 2}^4 \left( {{x^3} + {y^2} + z} \right){\rm{d}}x{\rm{d}}y{\rm{d}}z = 693$.
