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

\(\displaystyle \begin{align*}&\int_{3.5}^{8.5}\int_{-5}^{-1.5}\int_{-4}^{2}(\frac{(58e^{(x)})}{(7\cdot 2^y\cdot 3^{(\frac{z}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{8.5}\int_{-5}^{-1.5}\int_{-4}^{2}(\frac{(58e^{(x)})}{(7\cdot 2^y\cdot 3^{(\frac{z}{2})})})dxdydz=\int_{3.5}^{8.5}\int_{-5}^{-1.5}(\frac{(58e^{(x)})}{(7\cdot 2^y\cdot 3^{(\frac{z}{2})})})dydz|_{-4}^{2}\\&\int_{3.5}^{8.5}\int_{-5}^{-1.5}(\frac{(58\cdot (e^{(2)} - e^{(-4)}))}{(7\cdot 2^y\cdot 3^{(\frac{z}{2})})})dydz=\int_{3.5}^{8.5}(-\frac{(58e^{(-4)}\cdot (e^{(6)} - 1))}{(7\cdot 2^y\cdot 3^{(\frac{z}{2})}ln(2))})dz|_{-5}^{-1.5}\\&\int_{3.5}^{8.5}(\frac{(1856e^{(2)} - 1856e^{(-4)} - 116\cdot 2^{(\frac{1}{2})}e^{(2)} + 116\cdot 2^{(\frac{1}{2})}e^{(-4)})}{(7\cdot 3^{(\frac{z}{2})}ln(2))})dz=-\frac{(232e^{(-4)}\cdot (16e^{(6)} - 2^{(\frac{1}{2})}e^{(6)} + 2^{(\frac{1}{2})} - 16))}{(7\cdot 3^{(\frac{z}{2})}ln(2)ln(3))}|_{3.5}^{8.5}=640.3\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{5}^{10}\int_{-6}^{-4}\int_{7}^{8.5}(\frac{(2\cdot 3^z)}{(7xy)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{5}^{10}\int_{-6}^{-4}\int_{7}^{8.5}(\frac{(2\cdot 3^z)}{(7xy)})dxdydz=\int_{5}^{10}\int_{-6}^{-4}(\frac{(2\cdot 3^zln(x))}{(7y)})dydz|_{7}^{8.5}\\&\int_{5}^{10}\int_{-6}^{-4}(\frac{(2\cdot 3^zln(\frac{17}{14}))}{(7y)})dydz=\int_{5}^{10}(\frac{(2\cdot 3^zln(\frac{17}{14})ln(y))}{7})dz|_{-6}^{-4}\\&\int_{5}^{10}(\frac{(2\cdot 3^zln(\frac{2}{3})ln(\frac{17}{14}))}{7})dz=\frac{(2\cdot 3^zln(\frac{2}{3})ln(\frac{17}{14}))}{(7ln(3))}|_{5}^{10}=-1204\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-4.5}^{-3.5}\int_{6}^{10}\int_{10}^{15}(\frac{(26cos(y + 2))}{(9\cdot 3^z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-4.5}^{-3.5}\int_{6}^{10}\int_{10}^{15}(\frac{(26cos(y + 2))}{(9\cdot 3^z)})dxdydz=\int_{-4.5}^{-3.5}\int_{6}^{10}(\frac{(26xcos(y + 2))}{(9\cdot 3^z)})dydz|_{10}^{15}\\&\int_{-4.5}^{-3.5}\int_{6}^{10}(\frac{(130cos(y + 2))}{(9\cdot 3^z)})dydz=\int_{-4.5}^{-3.5}(\frac{(130sin(y + 2))}{(9\cdot 3^z)})dz|_{6}^{10}\\&\int_{-4.5}^{-3.5}(-\frac{(130\cdot (sin(8) - sin(12)))}{(9\cdot 3^z)})dz=\frac{(130\cdot (sin(8) - sin(12)))}{(9\cdot 3^zln(3))}|_{-4.5}^{-3.5}=-1876\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-8}^{-4}\int_{-7}^{-2.5}\int_{-4}^{-3}(\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-8}^{-4}\int_{-7}^{-2.5}\int_{-4}^{-3}(\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dxdydz=\int_{-8}^{-4}\int_{-7}^{-2.5}(-\frac{(3cos(y + 2)cos(z + 1)e^{(-x)})}{47})dydz|_{-4}^{-3}\\&\int_{-8}^{-4}\int_{-7}^{-2.5}(-\frac{(3cos(y + 2)cos(z + 1)\cdot (e^{(3)} - e^{(4)}))}{47})dydz=\int_{-8}^{-4}(-\frac{(3cos(z + 1)sin(y + 2)\cdot (e^{(3)} - e^{(4)}))}{47})dz|_{-7}^{-2.5}\\&\int_{-8}^{-4}(-\frac{(3cos(z + 1)e^{(3)}\cdot (sin(\frac{1}{2}) - sin(5))\cdot (e^{(1)} - 1))}{47})dz=-\frac{(3sin(z + 1)e^{(3)}\cdot (sin(\frac{1}{2}) - sin(5))\cdot (e^{(1)} - 1))}{47}|_{-8}^{-4}=-1.64\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{8}^{12}\int_{3}^{6}\int_{-6}^{-1.5}(\frac{(7z^{2}cos(3x)e^{(2y)})}{68})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{8}^{12}\int_{3}^{6}\int_{-6}^{-1.5}(\frac{(7z^{2}cos(3x)e^{(2y)})}{68})dxdydz=\int_{8}^{12}\int_{3}^{6}(\frac{(7z^{2}sin(3x)e^{(2y)})}{204})dydz|_{-6}^{-1.5}\\&\int_{8}^{12}\int_{3}^{6}(-\frac{(7z^{2}e^{(2y)}\cdot (\frac{sin(\frac{9}{2})}{3}-\frac{ sin(18)}{3}))}{68})dydz=\int_{8}^{12}(-\frac{(7z^{2}e^{(2y)}\cdot (\frac{sin(\frac{9}{2})}{3}-\frac{ sin(18)}{3}))}{136})dz|_{3}^{6}\\&\int_{8}^{12}(-\frac{(7z^{2}e^{(6)}\cdot (sin(\frac{9}{2}) - sin(18))\cdot (e^{(6)} - 1))}{408})dz=z^{3}\cdot (\frac{(7e^{(6)}sin(\frac{9}{2}))}{1224}-\frac{ (7e^{(12)}sin(\frac{9}{2}))}{1224}-\frac{ (7e^{(6)}sin(18))}{1224}+\frac{ (7e^{(12)}sin(18))}{1224})|_{8}^{12}=255800\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-7}^{-2.5}\int_{-5}^{-1}\int_{-8}^{-6.5}(\frac{12}{(31\cdot 3^yz)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-2.5}\int_{-5}^{-1}\int_{-8}^{-6.5}(\frac{12}{(31\cdot 3^yz)})dxdydz=\int_{-7}^{-2.5}\int_{-5}^{-1}(\frac{(12x)}{(31\cdot 3^yz)})dydz|_{-8}^{-6.5}\\&\int_{-7}^{-2.5}\int_{-5}^{-1}(\frac{18}{(31\cdot 3^yz)})dydz=\int_{-7}^{-2.5}(-\frac{18}{(31\cdot 3^yzln(3))})dz|_{-5}^{-1}\\&\int_{-7}^{-2.5}(\frac{4320}{(31zln(3))})dz=\frac{(4320ln(z))}{(31ln(3))}|_{-7}^{-2.5}=-130.6\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{3.5}^{6}\int_{-7}^{-2.5}\int_{8}^{10}(\frac{(23e^{(z)})}{(3x^{2}y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{3.5}^{6}\int_{-7}^{-2.5}\int_{8}^{10}(\frac{(23e^{(z)})}{(3x^{2}y^{2})})dxdydz=\int_{3.5}^{6}\int_{-7}^{-2.5}(-\frac{(23e^{(z)})}{(3xy^{2})})dydz|_{8}^{10}\\&\int_{3.5}^{6}\int_{-7}^{-2.5}(\frac{(23e^{(z)})}{(120y^{2})})dydz=\int_{3.5}^{6}(-\frac{(23e^{(z)})}{(120y)})dz|_{-7}^{-2.5}\\&\int_{3.5}^{6}(\frac{(69e^{(z)})}{1400})dz=\frac{(69e^{(z)})}{1400}|_{3.5}^{6}=18.25\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-5}^{-1.5}\int_{-5}^{-4}\int_{8}^{12.5}(\frac{(19sin(4x)e^{(-z)}e^{(y)})}{4})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-5}^{-1.5}\int_{-5}^{-4}\int_{8}^{12.5}(\frac{(19sin(4x)e^{(-z)}e^{(y)})}{4})dxdydz=\int_{-5}^{-1.5}\int_{-5}^{-4}(-\frac{(19cos(4x)e^{(y - z)})}{16})dydz|_{8}^{12.5}\\&\int_{-5}^{-1.5}\int_{-5}^{-4}(e^{(y - z)}\cdot (\frac{(19cos(32))}{16}-\frac{ (19cos(50))}{16}))dydz=\int_{-5}^{-1.5}(\frac{(19e^{(y - z)}\cdot (cos(32) - cos(50)))}{16})dz|_{-5}^{-4}\\&\int_{-5}^{-1.5}(\frac{(19e^{(- z - 5)}\cdot (cos(32) - cos(50))\cdot (e^{(1)} - 1))}{16})dz=-\frac{(19e^{(- z - 5)}\cdot (cos(32) - cos(50))\cdot (e^{(1)} - 1))}{16}|_{-5}^{-1.5}=-0.26\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-5}^{-3.5}\int_{10}^{11.5}\int_{3.5}^{8.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^zsin(y + 2))}{8})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-5}^{-3.5}\int_{10}^{11.5}\int_{3.5}^{8.5}(\frac{(3^{(\frac{x}{3})}\cdot 3^zsin(y + 2))}{8})dxdydz=\int_{-5}^{-3.5}\int_{10}^{11.5}(\frac{(3\cdot 3^{(\frac{x}{3}+ z)}sin(y + 2))}{(8ln(3))})dydz|_{3.5}^{8.5}\end{align*}\)

\(\displaystyle \begin{align*}\\&\int_{-5}^{-3.5}\int_{10}^{11.5}(-\frac{(3^zsin(y + 2)\cdot (9\cdot 3^{(\frac{1}{6})} - 27\cdot 3^{(\frac{5}{6})}))}{(8ln(3))})dydz=\int_{-5}^{-3.5}(\frac{(3^zcos(y + 2)\cdot (9\cdot 3^{(\frac{1}{6})} - 27\cdot 3^{(\frac{5}{6})}))}{(8ln(3))})dz|_{10}^{11.5}\\&\int_{-5}^{-3.5}(\frac{(9\cdot 3^{(\frac{1}{6})}\cdot 3^z\cdot (3\cdot 3^{(\frac{2}{3})} - 1)\cdot (cos(12) - cos(\frac{27}{2})))}{(8ln(3))})dz=-\frac{(9\cdot 3^z\cdot (3^{(\frac{1}{6})}cos(12) - 3^{(\frac{1}{6})}cos(\frac{27}{2}) - 3\cdot 243^{(\frac{1}{6})}cos(12) + 3\cdot 243^{(\frac{1}{6})}cos(\frac{27}{2})))}{(8ln(3)^{2})}|_{-5}^{-3.5}=0.03\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-3}\int_{8}^{12}\int_{10}^{13.5}(\frac{(10cos(4y)sin(x + 2))}{(9z^{3})})dxdydz=\int_{-7}^{-3}\int_{8}^{12}(-\frac{(10cos(x + 2)cos(4y))}{(9z^{3})})dydz|_{10}^{13.5}\\&\int_{-7}^{-3}\int_{8}^{12}(\frac{(10cos(4y)\cdot (cos(12) - cos(\frac{31}{2})))}{(9z^{3})})dydz=\int_{-7}^{-3}(\frac{(5sin(4y)\cdot (cos(12) - cos(\frac{31}{2})))}{(18z^{3})})dz|_{8}^{12}\\&\int_{-7}^{-3}(-\frac{(5\cdot (cos(12) - cos(\frac{31}{2}))\cdot (sin(32) - sin(48)))}{(18z^{3})})dz=\frac{(\frac{(5cos(12)sin(32))}{36}-\frac{ (5cos(12)sin(48))}{36}-\frac{ (5cos(\frac{31}{2})sin(32))}{36}+\frac{ (5cos(\frac{31}{2})sin(48))}{36})}{z^{2}}|_{-7}^{-3}=0.03\end{align*}\)