$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\end{align*}$

$\displaystyle \begin{align*}&\int_{-8}^{-3.5}\int_{-4.5}^{-0.5}\int_{4.5}^{7.5}(\frac{(45cos(z + 2)e^{(2x)})}{(11\cdot 2^y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-3.5}\int_{-4.5}^{-0.5}\int_{4.5}^{7.5}(\frac{(45cos(z + 2)e^{(2x)})}{(11\cdot 2^y)})dxdydz=\int_{-8}^{-3.5}\int_{-4.5}^{-0.5}(\frac{(45cos(z + 2)e^{(2x)})}{(22\cdot 2^y)})dydz|_{4.5}^{7.5}\\&\int_{-8}^{-3.5}\int_{-4.5}^{-0.5}(\frac{(45cos(z + 2)e^{(9)}\cdot (e^{(6)} - 1))}{(22\cdot 2^y)})dydz=\int_{-8}^{-3.5}(-\frac{(45cos(z + 2)e^{(9)}\cdot (e^{(6)} - 1))}{(22\cdot 2^yln(2))})dz|_{-4.5}^{-0.5}\\&\int_{-8}^{-3.5}(-\frac{(675\cdot 2^{(\frac{1}{2})}cos(z + 2)e^{(9)} - 675\cdot 2^{(\frac{1}{2})}cos(z + 2)e^{(15)})}{(22ln(2))})dz=\frac{(675\cdot 2^{(\frac{1}{2})}sin(z + 2)e^{(9)}\cdot (e^{(6)} - 1))}{(22ln(2))}|_{-8}^{-3.5}=-2.607\cdot10^{8}\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\end{align*}$

$\displaystyle \begin{align*}&\int_{7}^{9.5}\int_{-8}^{-7}\int_{-3}^{0.5}(\frac{(9sin(4y)e^{(x)})}{35})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{9.5}\int_{-8}^{-7}\int_{-3}^{0.5}(\frac{(9sin(4y)e^{(x)})}{35})dxdydz=\int_{7}^{9.5}\int_{-8}^{-7}(\frac{(9sin(4y)e^{(x)})}{35})dydz|_{-3}^{0.5}\\&\int_{7}^{9.5}\int_{-8}^{-7}(-\frac{(9sin(4y)\cdot (e^{(-3)} - e^{(\frac{1}{2})}))}{35})dydz=\int_{7}^{9.5}(-\frac{(9cos(4y)e^{(-3)}\cdot (e^{(\frac{7}{2})} - 1))}{140})dz|_{-8}^{-7}\\&\int_{7}^{9.5}(-\frac{(9e^{(-3)}\cdot (cos(28) - cos(32))\cdot (e^{(\frac{7}{2})} - 1))}{140})dz=-\frac{(9ze^{(-3)}\cdot (cos(28) - cos(32))\cdot (e^{(\frac{7}{2})} - 1))}{140}|_{7}^{9.5}=0.46\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\end{align*}$

$\displaystyle \begin{align*}&\int_{6}^{9}\int_{-8}^{-4.5}\int_{3.5}^{7.5}(\frac{(7ze^{(2x)})}{(4y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{9}\int_{-8}^{-4.5}\int_{3.5}^{7.5}(\frac{(7ze^{(2x)})}{(4y^{2})})dxdydz=\int_{6}^{9}\int_{-8}^{-4.5}(\frac{(7ze^{(2x)})}{(8y^{2})})dydz|_{3.5}^{7.5}\\&\int_{6}^{9}\int_{-8}^{-4.5}(\frac{(7ze^{(7)}\cdot (e^{(8)} - 1))}{(8y^{2})})dydz=\int_{6}^{9}(-\frac{(7ze^{(7)}\cdot (e^{(8)} - 1))}{(8y)})dz|_{-8}^{-4.5}\\&\int_{6}^{9}(\frac{(49ze^{(7)}\cdot (e^{(8)} - 1))}{576})dz=\frac{(49z^{2}e^{(7)}\cdot (e^{(8)} - 1))}{1152}|_{6}^{9}=6.255\cdot10^{6}\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\end{align*}$

$\displaystyle \begin{align*}&\int_{10}^{13}\int_{-6}^{-3.5}\int_{-4.5}^{-1}(\frac{e^{(-x)}}{(21yz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{13}\int_{-6}^{-3.5}\int_{-4.5}^{-1}(\frac{e^{(-x)}}{(21yz)})dxdydz=\int_{10}^{13}\int_{-6}^{-3.5}(-\frac{e^{(-x)}}{(21yz)})dydz|_{-4.5}^{-1}\\&\int_{10}^{13}\int_{-6}^{-3.5}(-\frac{(e^{(1)} - e^{(\frac{9}{2})})}{(21yz)})dydz=\int_{10}^{13}(\frac{(e^{(1)}ln(y)\cdot (e^{(\frac{7}{2})} - 1))}{(21z)})dz|_{-6}^{-3.5}\\&\int_{10}^{13}(\frac{(e^{(1)}ln(\frac{7}{12})\cdot (e^{(\frac{7}{2})} - 1))}{(21z)})dz=\frac{(e^{(1)}ln(\frac{7}{12})ln(z)\cdot (e^{(\frac{7}{2})} - 1))}{21}|_{10}^{13}=-0.59\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\end{align*}$

$\displaystyle \begin{align*}&\int_{10}^{12}\int_{3.5}^{6.5}\int_{-10}^{-5}(\frac{(8cos(3x)e^{(y)})}{z})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{12}\int_{3.5}^{6.5}\int_{-10}^{-5}(\frac{(8cos(3x)e^{(y)})}{z})dxdydz=\int_{10}^{12}\int_{3.5}^{6.5}(\frac{(8sin(3x)e^{(y)})}{(3z)})dydz|_{-10}^{-5}\\&\int_{10}^{12}\int_{3.5}^{6.5}(-\frac{(8e^{(y)}\cdot (sin(15) - sin(30)))}{(3z)})dydz=\int_{10}^{12}(-\frac{(8e^{(y)}\cdot (sin(15) - sin(30)))}{(3z)})dz|_{3.5}^{6.5}\\&\int_{10}^{12}(-\frac{(8e^{(\frac{7}{2})}\cdot (sin(15) - sin(30))\cdot (e^{(3)} - 1))}{(3z)})dz=ln(z)\cdot (\frac{(8e^{(\frac{7}{2})}sin(15))}{3}-\frac{ (8e^{(\frac{13}{2})}sin(15))}{3}-\frac{ (8e^{(\frac{7}{2})}sin(30))}{3}+\frac{ (8e^{(\frac{13}{2})}sin(30))}{3})|_{10}^{12}=-503.4\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\displaystyle \begin{align*}&\int_{9}^{12}\int_{3.5}^{4.5}\int_{4.5}^{8}(\frac{(12\cdot 3^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})})}{(7z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{12}\int_{3.5}^{4.5}\int_{4.5}^{8}(\frac{(12\cdot 3^{(\frac{x}{4})}\cdot 3^{(\frac{y}{2})})}{(7z^{2})})dxdydz=\int_{9}^{12}\int_{3.5}^{4.5}(\frac{(48\cdot 3^{(\frac{x}{4}+\frac{ y}{2})})}{(7z^{2}ln(3))})dydz|_{4.5}^{8}\\&\int_{9}^{12}\int_{3.5}^{4.5}(-\frac{(144\cdot 3^{(\frac{y}{2})}\cdot (3^{(\frac{1}{8})} - 3))}{(7z^{2}ln(3))})dydz=\int_{9}^{12}(-\frac{(288\cdot 3^{(\frac{y}{2})}\cdot (3^{(\frac{1}{8})} - 3))}{(7z^{2}ln(3)^{2})})dz|_{3.5}^{4.5}\\&\int_{9}^{12}(\frac{(7776\cdot 3^{(\frac{1}{4})} - 2592\cdot 3^{(\frac{3}{4})} - 2592\cdot 3^{(\frac{3}{8})} + 864\cdot 3^{(\frac{7}{8})})}{(7z^{2}ln(3)^{2})})dz=-\frac{(7776\cdot 3^{(\frac{1}{4})} - 2592\cdot 27^{(\frac{1}{4})} - 2592\cdot 27^{(\frac{1}{8})} + 864\cdot 2187^{(\frac{1}{8})})}{(7zln(3)^{2})}|_{9}^{12}=8.78\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\end{align*}$

$\displaystyle \begin{align*}&\int_{10}^{15}\int_{5}^{9.5}\int_{6}^{7.5}(\frac{(71e^{(y)})}{(6xz^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{15}\int_{5}^{9.5}\int_{6}^{7.5}(\frac{(71e^{(y)})}{(6xz^{3})})dxdydz=\int_{10}^{15}\int_{5}^{9.5}(\frac{(71e^{(y)}ln(x))}{(6z^{3})})dydz|_{6}^{7.5}\\&\int_{10}^{15}\int_{5}^{9.5}(\frac{(71e^{(y)}ln(\frac{5}{4}))}{(6z^{3})})dydz=\int_{10}^{15}(\frac{(71e^{(y)}ln(\frac{5}{4}))}{(6z^{3})})dz|_{5}^{9.5}\\&\int_{10}^{15}(\frac{(71e^{(5)}ln(\frac{5}{4})\cdot (e^{(\frac{9}{2})} - 1))}{(6z^{3})})dz=-\frac{(71e^{(5)}ln(\frac{5}{4})\cdot (e^{(\frac{9}{2})} - 1))}{(12z^{2})}|_{10}^{15}=96.9\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\displaystyle \begin{align*}&\int_{-7}^{-4.5}\int_{8}^{9.5}\int_{-4.5}^{-2.5}(\frac{29}{(2\cdot 2^xyz^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-4.5}\int_{8}^{9.5}\int_{-4.5}^{-2.5}(\frac{29}{(2\cdot 2^xyz^{3})})dxdydz=\int_{-7}^{-4.5}\int_{8}^{9.5}(-\frac{29}{(2\cdot 2^xyz^{3}ln(2))})dydz|_{-4.5}^{-2.5}\\&\int_{-7}^{-4.5}\int_{8}^{9.5}(\frac{(174\cdot 2^{(\frac{1}{2})})}{(yz^{3}ln(2))})dydz=\int_{-7}^{-4.5}(\frac{(174\cdot 2^{(\frac{1}{2})}ln(y))}{(z^{3}ln(2))})dz|_{8}^{9.5}\\&\int_{-7}^{-4.5}(\frac{(174\cdot 2^{(\frac{1}{2})}ln(\frac{19}{16}))}{(z^{3}ln(2))})dz=-\frac{(87\cdot 2^{(\frac{1}{2})}ln(\frac{19}{16}))}{(z^{2}ln(2))}|_{-7}^{-4.5}=-0.88\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\displaystyle \begin{align*}&\int_{-8}^{-6.5}\int_{-3.5}^{0.5}\int_{-10}^{-5}(\frac{(76\cdot 3^y)}{(27xz^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-6.5}\int_{-3.5}^{0.5}\int_{-10}^{-5}(\frac{(76\cdot 3^y)}{(27xz^{2})})dxdydz=\int_{-8}^{-6.5}\int_{-3.5}^{0.5}(\frac{(76\cdot 3^yln(x))}{(27z^{2})})dydz|_{-10}^{-5}\\&\int_{-8}^{-6.5}\int_{-3.5}^{0.5}(-\frac{(76\cdot 3^yln(2))}{(27z^{2})})dydz=\int_{-8}^{-6.5}(-\frac{(76\cdot 3^yln(2))}{(27z^{2}ln(3))})dz|_{-3.5}^{0.5}\\&\int_{-8}^{-6.5}(-\frac{(6080\cdot 3^{(\frac{1}{2})}ln(2))}{(2187z^{2}ln(3))})dz=\frac{(6080\cdot 3^{(\frac{1}{2})}ln(2))}{(2187zln(3))}|_{-8}^{-6.5}=-0.09\end{align*}$

$\displaystyle \begin{align*}&\text{Performing a triple integral, the order of integration does}\\&\text{not entirely matter.}\\&\text{For example:}\\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\\&\text{Considering our integral, and keeping savvy about }\\&\text{utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\displaystyle \begin{align*}&\int_{-7}^{-4}\int_{3}^{8}\int_{3}^{6.5}(\frac{(11\cdot 3^{(\frac{y}{2})}e^{(-x)})}{(180z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-4}\int_{3}^{8}\int_{3}^{6.5}(\frac{(11\cdot 3^{(\frac{y}{2})}e^{(-x)})}{(180z)})dxdydz=\int_{-7}^{-4}\int_{3}^{8}(-\frac{(11\cdot 3^{(\frac{y}{2})}e^{(-x)})}{(180z)})dydz|_{3}^{6.5}\\&\int_{-7}^{-4}\int_{3}^{8}(\frac{(11\cdot 3^{(\frac{y}{2})}\cdot (e^{(-3)} - e^{(-\frac{13}{2})}))}{(180z)})dydz=\int_{-7}^{-4}(\frac{(11\cdot 3^{(\frac{y}{2})}e^{(-\frac{13}{2})}\cdot (e^{(\frac{7}{2})} - 1))}{(90zln(3))})dz|_{3}^{8}\\&\int_{-7}^{-4}(\frac{(297e^{(-3)} - 297e^{(-\frac{13}{2})} - 11\cdot 3^{(\frac{1}{2})}e^{(-3)} + 11\cdot 3^{(\frac{1}{2})}e^{(-\frac{13}{2})})}{(30zln(3))})dz=\frac{(e^{(-\frac{13}{2})}ln(z)\cdot (297e^{(\frac{7}{2})} - 11\cdot 3^{(\frac{1}{2})}e^{(\frac{7}{2})} + 11\cdot 3^{(\frac{1}{2})} - 297))}{(30ln(3))}|_{-7}^{-4}=-0.23\end{align*}$