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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\displaystyle \begin{align*}&\int_{9}^{13.5}\int_{3}^{7.5}\int_{2}^{5}(\frac{(5sin(z + 2))}{(8\cdot 3^{(\frac{y}{2})}x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{9}^{13.5}\int_{3}^{7.5}\int_{2}^{5}(\frac{(5sin(z + 2))}{(8\cdot 3^{(\frac{y}{2})}x)})dxdydz=\int_{9}^{13.5}\int_{3}^{7.5}(\frac{(5sin(z + 2)ln(x))}{(8\cdot 3^{(\frac{y}{2})})})dydz|_{2}^{5}\\&\int_{9}^{13.5}\int_{3}^{7.5}(\frac{(5sin(z + 2)ln(\frac{5}{2}))}{(8\cdot 3^{(\frac{y}{2})})})dydz=\int_{9}^{13.5}(-\frac{(5sin(z + 2)ln(\frac{5}{2}))}{(4\cdot 3^{(\frac{y}{2})}ln(3))})dz|_{3}^{7.5}\\&\int_{9}^{13.5}(\frac{(5sin(z + 2)ln(\frac{5}{2})\cdot (18\cdot 3^{(\frac{1}{2})} - 2\cdot 3^{(\frac{1}{4})}))}{(648ln(3))})dz=-\frac{(5cos(z + 2)ln(\frac{5}{2})\cdot (18\cdot 3^{(\frac{1}{2})} - 2\cdot 3^{(\frac{1}{4})}))}{(648ln(3))}|_{9}^{13.5}=0.18\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-7}^{-4.5}\int_{-3}^{1.5}\int_{3}^{5}(\frac{(11e^{(-2y)})}{(38\cdot 2^x)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-4.5}\int_{-3}^{1.5}\int_{3}^{5}(\frac{(11e^{(-2y)})}{(38\cdot 2^x)})dxdydz=\int_{-7}^{-4.5}\int_{-3}^{1.5}(-\frac{(11e^{(-2y)})}{(38\cdot 2^xln(2))})dydz|_{3}^{5}\\&\int_{-7}^{-4.5}\int_{-3}^{1.5}(\frac{(33e^{(-2y)})}{(1216ln(2))})dydz=\int_{-7}^{-4.5}(-\frac{(33e^{(-2y)})}{(2432ln(2))})dz|_{-3}^{1.5}\\&\int_{-7}^{-4.5}(\frac{(33e^{(-3)}\cdot (e^{(9)} - 1))}{(2432ln(2))})dz=\frac{(33ze^{(-3)}\cdot (e^{(9)} - 1))}{(2432ln(2))}|_{-7}^{-4.5}=19.74\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-5}^{-4}\int_{7}^{12}\int_{6}^{10.5}(\frac{(11sin(x + 1)e^{(-z)})}{(31y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-5}^{-4}\int_{7}^{12}\int_{6}^{10.5}(\frac{(11sin(x + 1)e^{(-z)})}{(31y^{2})})dxdydz=\int_{-5}^{-4}\int_{7}^{12}(-\frac{(11cos(x + 1)e^{(-z)})}{(31y^{2})})dydz|_{6}^{10.5}\\&\int_{-5}^{-4}\int_{7}^{12}(\frac{(11e^{(-z)}\cdot (cos(7) - cos(\frac{23}{2})))}{(31y^{2})})dydz=\int_{-5}^{-4}(-\frac{(11e^{(-z)}\cdot (cos(7) - cos(\frac{23}{2})))}{(31y)})dz|_{7}^{12}\\&\int_{-5}^{-4}(\frac{(55e^{(-z)}\cdot (cos(7) - cos(\frac{23}{2})))}{2604})dz=-\frac{(55e^{(-z)}\cdot (cos(7) - cos(\frac{23}{2})))}{2604}|_{-5}^{-4}=0.54\end{align*}\)