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

\(\displaystyle \begin{align*}&\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{10}\int_{9}^{13.5}\int_{-3.5}^{-1.5}(\frac{(17\cdot 3^{(\frac{x}{3})}sin(y + 2))}{10})dxdydz=\int_{8}^{10}\int_{9}^{13.5}(\frac{(51\cdot 3^{(\frac{x}{3})}sin(y + 2))}{(10ln(3))})dydz|_{-3.5}^{-1.5}\\&\int_{8}^{10}\int_{9}^{13.5}(\frac{(17sin(y + 2)\cdot (3\cdot 3^{(\frac{1}{2})} - 3^{(\frac{5}{6})}))}{(30ln(3))})dydz=\int_{8}^{10}(-\frac{(17cos(y + 2)\cdot (3\cdot 3^{(\frac{1}{2})} - 3^{(\frac{5}{6})}))}{(30ln(3))})dz|_{9}^{13.5}\\&\int_{8}^{10}(-\frac{(17\cdot 3^{(\frac{1}{2})}\cdot (cos(11) - cos(\frac{31}{2}))\cdot (3^{(\frac{1}{3})} - 3))}{(30ln(3))})dz=-\frac{(17\cdot 3^{(\frac{1}{2})}z\cdot (cos(11) - cos(\frac{31}{2}))\cdot (3^{(\frac{1}{3})} - 3))}{(30ln(3))}|_{8}^{10}=2.74\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{8}^{11.5}\int_{-10}^{-5}\int_{-9}^{-4}(\frac{(3cos(x + 1)cos(z + 2))}{(23y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{11.5}\int_{-10}^{-5}\int_{-9}^{-4}(\frac{(3cos(x + 1)cos(z + 2))}{(23y)})dxdydz=\int_{8}^{11.5}\int_{-10}^{-5}(\frac{(3cos(z + 2)sin(x + 1))}{(23y)})dydz|_{-9}^{-4}\\&\int_{8}^{11.5}\int_{-10}^{-5}(-\frac{(3cos(z + 2)\cdot (sin(3) - sin(8)))}{(23y)})dydz=\int_{8}^{11.5}(-ln(y)\cdot (\frac{(3cos(z + 2)sin(3))}{23}-\frac{ (3cos(z + 2)sin(8))}{23}))dz|_{-10}^{-5}\\&\int_{8}^{11.5}(\frac{(3cos(z + 2)ln(2)\cdot (sin(3) - sin(8)))}{23})dz=\frac{(3sin(z + 2)ln(2)\cdot (sin(3) - sin(8)))}{23}|_{8}^{11.5}=-0.1\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-8}^{-4.5}\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(61sin(z + 1))}{(33\cdot 3^xy)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-4.5}\int_{-6}^{-4}\int_{-3}^{-1}(\frac{(61sin(z + 1))}{(33\cdot 3^xy)})dxdydz=\int_{-8}^{-4.5}\int_{-6}^{-4}(-\frac{(61sin(z + 1))}{(33\cdot 3^xyln(3))})dydz|_{-3}^{-1}\\&\int_{-8}^{-4.5}\int_{-6}^{-4}(\frac{(488sin(z + 1))}{(11yln(3))})dydz=\int_{-8}^{-4.5}(\frac{(488sin(z + 1)ln(y))}{(11ln(3))})dz|_{-6}^{-4}\\&\int_{-8}^{-4.5}(\frac{(488sin(z + 1)ln(\frac{2}{3}))}{(11ln(3))})dz=-\frac{(488cos(z + 1)ln(\frac{2}{3}))}{(11ln(3))}|_{-8}^{-4.5}=-27.68\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-4.5}^{-2}\int_{-3}^{-1.5}\int_{3.5}^{6.5}(\frac{(49\cdot 2^{(\frac{x}{4})}e^{(2y)}e^{(-z)})}{2})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-2}\int_{-3}^{-1.5}\int_{3.5}^{6.5}(\frac{(49\cdot 2^{(\frac{x}{4})}e^{(2y)}e^{(-z)})}{2})dxdydz=\int_{-4.5}^{-2}\int_{-3}^{-1.5}(\frac{(98\cdot 2^{(\frac{x}{4})}e^{(2y - z)})}{ln(2)})dydz|_{3.5}^{6.5}\\&\int_{-4.5}^{-2}\int_{-3}^{-1.5}(-\frac{(98\cdot 2^{(\frac{5}{8})}e^{(2y - z)}\cdot (2^{(\frac{1}{4})} - 2))}{ln(2)})dydz=\int_{-4.5}^{-2}(-\frac{(49\cdot 2^{(\frac{5}{8})}e^{(2y - z)}\cdot (2^{(\frac{1}{4})} - 2))}{ln(2)})dz|_{-3}^{-1.5}\\&\int_{-4.5}^{-2}(-\frac{(49\cdot 2^{(\frac{5}{8})}e^{(- z - 6)}\cdot (2^{(\frac{1}{4})} - 2)\cdot (e^{(3)} - 1))}{ln(2)})dz=\frac{(49\cdot 2^{(\frac{5}{8})}e^{(- z - 6)}\cdot (2^{(\frac{1}{4})} - 2)\cdot (e^{(3)} - 1))}{ln(2)}|_{-4.5}^{-2}=345.5\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{6}^{11}\int_{10}^{11.5}\int_{9}^{13.5}(\frac{(25ysin(x + 2))}{(2z^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{11}\int_{10}^{11.5}\int_{9}^{13.5}(\frac{(25ysin(x + 2))}{(2z^{3})})dxdydz=\int_{6}^{11}\int_{10}^{11.5}(-\frac{(25ycos(x + 2))}{(2z^{3})})dydz|_{9}^{13.5}\\&\int_{6}^{11}\int_{10}^{11.5}(\frac{(25y\cdot (cos(11) - cos(\frac{31}{2})))}{(2z^{3})})dydz=\int_{6}^{11}(\frac{(y^{2}\cdot (25cos(11) - 25cos(\frac{31}{2})))}{(4z^{3})})dz|_{10}^{11.5}\\&\int_{6}^{11}(\frac{(3225\cdot (cos(11) - cos(\frac{31}{2})))}{(16z^{3})})dz=-\frac{(\frac{(3225cos(11))}{32}-\frac{ (3225cos(\frac{31}{2}))}{32})}{z^{2}}|_{6}^{11}=1.93\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{7}^{11.5}\int_{-3}^{-2}\int_{-8}^{-6}(\frac{(49cos(3z)e^{(-y)})}{(12x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{11.5}\int_{-3}^{-2}\int_{-8}^{-6}(\frac{(49cos(3z)e^{(-y)})}{(12x)})dxdydz=\int_{7}^{11.5}\int_{-3}^{-2}(\frac{(49cos(3z)e^{(-y)}ln(x))}{12})dydz|_{-8}^{-6}\\&\int_{7}^{11.5}\int_{-3}^{-2}(\frac{(49cos(3z)e^{(-y)}ln(\frac{3}{4}))}{12})dydz=\int_{7}^{11.5}(-\frac{(49cos(3z)e^{(-y)}ln(\frac{3}{4}))}{12})dz|_{-3}^{-2}\\&\int_{7}^{11.5}(-\frac{(49cos(3z)ln(\frac{3}{4})\cdot (e^{(2)} - e^{(3)}))}{12})dz=-\frac{(49sin(3z)ln(\frac{3}{4})\cdot (e^{(2)} - e^{(3)}))}{36}|_{7}^{11.5}=3.87\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-6}^{-3}\int_{4}^{5.5}\int_{-3}^{-1}(\frac{(7z)}{(2^x\cdot 2^{(\frac{y}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-6}^{-3}\int_{4}^{5.5}\int_{-3}^{-1}(\frac{(7z)}{(2^x\cdot 2^{(\frac{y}{2})})})dxdydz=\int_{-6}^{-3}\int_{4}^{5.5}(-\frac{(\frac{7}{2^{(x +\frac{ y}{2})}z})}{ln(2)})dydz|_{-3}^{-1}\\&\int_{-6}^{-3}\int_{4}^{5.5}(\frac{(42z)}{(2^{(\frac{y}{2})}ln(2))})dydz=\int_{-6}^{-3}(-\frac{(84z)}{(2^{(\frac{y}{2})}ln(2)^{2})})dz|_{4}^{5.5}\\&\int_{-6}^{-3}(-\frac{(21z\cdot (2^{(\frac{1}{4})} - 2))}{(2ln(2)^{2})})dz=-\frac{(z^{2}\cdot (21\cdot 2^{(\frac{1}{4})} - 42))}{(4ln(2)^{2})}|_{-6}^{-3}=-239.2\end{align*}\)

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

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

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

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

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

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