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

\(\displaystyle \begin{align*}&\int_{5}^{6.5}\int_{6}^{7}\int_{-8}^{-7}(\frac{(12\cdot 3^zcos(3x))}{(47y^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{5}^{6.5}\int_{6}^{7}\int_{-8}^{-7}(\frac{(12\cdot 3^zcos(3x))}{(47y^{2})})dxdydz=\int_{5}^{6.5}\int_{6}^{7}(\frac{(4\cdot 3^zsin(3x))}{(47y^{2})})dydz|_{-8}^{-7}\\&\int_{5}^{6.5}\int_{6}^{7}(-\frac{(12\cdot 3^z\cdot (\frac{sin(21)}{3}-\frac{ sin(24)}{3}))}{(47y^{2})})dydz=\int_{5}^{6.5}(\frac{(\frac{(4\cdot 3^zsin(21))}{47}-\frac{ (4\cdot 3^zsin(24))}{47})}{y})dz|_{6}^{7}\\&\int_{5}^{6.5}(-\frac{(2\cdot 3^z\cdot (sin(21) - sin(24)))}{987})dz=-\frac{(2\cdot 3^z\cdot (sin(21) - sin(24)))}{(987ln(3))}|_{5}^{6.5}=-3.28\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{7}^{10}\int_{8}^{9.5}\int_{-7}^{-5}(\frac{(71cos(z + 1))}{(33y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{7}^{10}\int_{8}^{9.5}\int_{-7}^{-5}(\frac{(71cos(z + 1))}{(33y)})dxdydz=\int_{7}^{10}\int_{8}^{9.5}(\frac{(71xcos(z + 1))}{(33y)})dydz|_{-7}^{-5}\\&\int_{7}^{10}\int_{8}^{9.5}(\frac{(142cos(z + 1))}{(33y)})dydz=\int_{7}^{10}(\frac{(142cos(z + 1)ln(y))}{33})dz|_{8}^{9.5}\\&\int_{7}^{10}(\frac{(142cos(z + 1)ln(\frac{19}{16}))}{33})dz=\frac{(142sin(z + 1)ln(\frac{19}{16}))}{33}|_{7}^{10}=-1.47\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}\int_{9}^{10.5}(\frac{(3sin(x + 1))}{(35\cdot 2^y\cdot 3^z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}\int_{9}^{10.5}(\frac{(3sin(x + 1))}{(35\cdot 2^y\cdot 3^z)})dxdydz=\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}(-\frac{(3cos(x + 1))}{(35\cdot 2^y\cdot 3^z)})dydz|_{9}^{10.5}\\&\int_{-3.5}^{-1}\int_{-3.5}^{-2.5}(\frac{(3\cdot (cos(10) - cos(\frac{23}{2})))}{(35\cdot 2^y\cdot 3^z)})dydz=\int_{-3.5}^{-1}(-\frac{(3\cdot (\frac{1}{3})^z\cdot (cos(10) - cos(\frac{23}{2})))}{(35\cdot 2^yln(2))})dz|_{-3.5}^{-2.5}\\&\int_{-3.5}^{-1}(\frac{(12\cdot 2^{(\frac{1}{2})}cos(10) - 12\cdot 2^{(\frac{1}{2})}cos(\frac{23}{2}))}{(35\cdot 3^zln(2))})dz=-\frac{(12\cdot (2^{(\frac{1}{2})}cos(10) - 2^{(\frac{1}{2})}cos(\frac{23}{2})))}{(35\cdot 3^zln(2)ln(3))}|_{-3.5}^{-1}=-36.85\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-3}^{-2}\int_{3}^{5}\int_{10}^{11.5}(\frac{(3e^{(2z)})}{(140\cdot 2^yx)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-3}^{-2}\int_{3}^{5}\int_{10}^{11.5}(\frac{(3e^{(2z)})}{(140\cdot 2^yx)})dxdydz=\int_{-3}^{-2}\int_{3}^{5}(\frac{(3e^{(2z)}ln(x))}{(140\cdot 2^y)})dydz|_{10}^{11.5}\\&\int_{-3}^{-2}\int_{3}^{5}(\frac{(3e^{(2z)}ln(\frac{23}{20}))}{(140\cdot 2^y)})dydz=\int_{-3}^{-2}(-\frac{(3e^{(2z)}ln(\frac{23}{20}))}{(140\cdot 2^yln(2))})dz|_{3}^{5}\\&\int_{-3}^{-2}(\frac{(9e^{(2z)}ln(\frac{23}{20}))}{(4480ln(2))})dz=\frac{(9e^{(2z)}ln(\frac{23}{20}))}{(8960ln(2))}|_{-3}^{-2}=3.21\cdot10^{-6}\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{-7}^{-3.5}\int_{-8}^{-5}\int_{10}^{12.5}(\frac{(7cos(y + 2)sin(x + 1)sin(3z))}{67})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{-7}^{-3.5}\int_{-8}^{-5}\int_{10}^{12.5}(\frac{(7cos(y + 2)sin(x + 1)sin(3z))}{67})dxdydz=\int_{-7}^{-3.5}\int_{-8}^{-5}(-\frac{(7cos(x + 1)cos(y + 2)sin(3z))}{67})dydz|_{10}^{12.5}\\&\int_{-7}^{-3.5}\int_{-8}^{-5}(\frac{(7cos(y + 2)sin(3z)\cdot (cos(11) - cos(\frac{27}{2})))}{67})dydz=\int_{-7}^{-3.5}(\frac{(7sin(y + 2)sin(3z)\cdot (cos(11) - cos(\frac{27}{2})))}{67})dz|_{-8}^{-5}\\&\int_{-7}^{-3.5}(-\frac{(7sin(3z)\cdot (cos(11) - cos(\frac{27}{2}))\cdot (sin(3) - sin(6)))}{67})dz=\frac{(7cos(3z)\cdot (cos(11) - cos(\frac{27}{2}))\cdot (sin(3) - sin(6)))}{201}|_{-7}^{-3.5}=-6.24\cdot10^{-4}\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{10}^{14}\int_{-8}^{-3}\int_{-7}^{-2}(\frac{sin(y + 2)}{(21x^{3}z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{14}\int_{-8}^{-3}\int_{-7}^{-2}(\frac{sin(y + 2)}{(21x^{3}z)})dxdydz=\int_{10}^{14}\int_{-8}^{-3}(-\frac{sin(y + 2)}{(42x^{2}z)})dydz|_{-7}^{-2}\\&\int_{10}^{14}\int_{-8}^{-3}(-\frac{(15sin(y + 2))}{(2744z)})dydz=\int_{10}^{14}(\frac{(15cos(y + 2))}{(2744z)})dz|_{-8}^{-3}\\&\int_{10}^{14}(\frac{(15\cdot (cos(1) - cos(6)))}{(2744z)})dz=ln(z)\cdot (\frac{(15cos(1))}{2744}-\frac{ (15cos(6))}{2744})|_{10}^{14}=-7.72\cdot10^{-4}\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{4.5}^{6}\int_{10}^{11}\int_{-7}^{-6}(\frac{(3^{(\frac{z}{2})}cos(4x))}{(11y)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{4.5}^{6}\int_{10}^{11}\int_{-7}^{-6}(\frac{(3^{(\frac{z}{2})}cos(4x))}{(11y)})dxdydz=\int_{4.5}^{6}\int_{10}^{11}(\frac{(3^{(\frac{z}{2})}sin(4x))}{(44y)})dydz|_{-7}^{-6}\\&\int_{4.5}^{6}\int_{10}^{11}(-\frac{(3^{(\frac{z}{2})}\cdot (\frac{sin(24)}{4}-\frac{ sin(28)}{4}))}{(11y)})dydz=\int_{4.5}^{6}(-\frac{(3^{(\frac{z}{2})}ln(y)\cdot (sin(24) - sin(28)))}{44})dz|_{10}^{11}\\&\int_{4.5}^{6}(-\frac{(3^{(\frac{z}{2})}ln(\frac{11}{10})\cdot (sin(24) - sin(28)))}{44})dz=-\frac{(3^{(\frac{z}{2})}\cdot (ln(\frac{11}{10})sin(24) - ln(\frac{11}{10})sin(28)))}{(22ln(3))}|_{4.5}^{6}=7.03\cdot10^{-2}\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{6}^{11}\int_{-3.5}^{0.5}\int_{-8}^{-3.5}(\frac{(5\cdot 2^{(\frac{y}{2})}cos(3x))}{(8z^{2})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{11}\int_{-3.5}^{0.5}\int_{-8}^{-3.5}(\frac{(5\cdot 2^{(\frac{y}{2})}cos(3x))}{(8z^{2})})dxdydz=\int_{6}^{11}\int_{-3.5}^{0.5}(\frac{(5\cdot 2^{(\frac{y}{2})}sin(3x))}{(24z^{2})})dydz|_{-8}^{-3.5}\\&\int_{6}^{11}\int_{-3.5}^{0.5}(-\frac{(5\cdot 2^{(\frac{y}{2})}\cdot (\frac{sin(\frac{21}{2})}{3}-\frac{ sin(24)}{3}))}{(8z^{2})})dydz=\int_{6}^{11}(-\frac{(5\cdot 2^{(\frac{y}{2})}\cdot (sin(\frac{21}{2}) - sin(24)))}{(12z^{2}ln(2))})dz|_{-3.5}^{0.5}\\&\int_{6}^{11}(-\frac{(5\cdot 2^{(\frac{1}{4})}sin(\frac{21}{2}) - 5\cdot 2^{(\frac{1}{4})}sin(24))}{(16z^{2}ln(2))})dz=\frac{(2^{(\frac{1}{4})}\cdot (\frac{(5sin(\frac{21}{2}))}{16}-\frac{ (5sin(24))}{16}))}{(zln(2))}|_{6}^{11}=-1.05\cdot10^{-3}\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{6}^{8}\int_{-9}^{-4}\int_{-10}^{-7}(\frac{1}{(x^{2}y^{2}z)})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{6}^{8}\int_{-9}^{-4}\int_{-10}^{-7}(\frac{1}{(x^{2}y^{2}z)})dxdydz=\int_{6}^{8}\int_{-9}^{-4}(-\frac{1}{(xy^{2}z)})dydz|_{-10}^{-7}\\&\int_{6}^{8}\int_{-9}^{-4}(\frac{3}{(70y^{2}z)})dydz=\int_{6}^{8}(-\frac{3}{(70yz)})dz|_{-9}^{-4}\\&\int_{6}^{8}(\frac{1}{(168z)})dz=\frac{ln(z)}{168}|_{6}^{8}=1.71\cdot10^{-3}\end{align*}\)

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

\(\displaystyle \begin{align*}&\int_{10}^{11}\int_{3}^{5.5}\int_{-3}^{-2}(\frac{(25\cdot 3^{(\frac{y}{2})}cos(z + 2))}{(7\cdot 3^{(\frac{x}{2})})})dxdydz\\&\text{The approach is simply to take it step by step:}\\&\int_{10}^{11}\int_{3}^{5.5}\int_{-3}^{-2}(\frac{(25\cdot 3^{(\frac{y}{2})}cos(z + 2))}{(7\cdot 3^{(\frac{x}{2})})})dxdydz=\int_{10}^{11}\int_{3}^{5.5}(-\frac{(50\cdot 3^{(\frac{y}{2}-\frac{ x}{2})}cos(z + 2))}{(7ln(3))})dydz|_{-3}^{-2}\\&\int_{10}^{11}\int_{3}^{5.5}(\frac{(150\cdot 3^{(\frac{y}{2})}cos(z + 2)\cdot (3^{(\frac{1}{2})} - 1))}{(7ln(3))})dydz=\int_{10}^{11}(-\frac{(300\cdot 3^{(\frac{y}{2})}\cdot (cos(z + 2) - 3^{(\frac{1}{2})}cos(z + 2)))}{(7ln(3)^{2})})dz|_{3}^{5.5}\\&\int_{10}^{11}(\frac{(900cos(z + 2)\cdot (3^{(\frac{1}{2})} + 9\cdot 3^{(\frac{1}{4})} - 3\cdot 3^{(\frac{3}{4})} - 3))}{(7ln(3)^{2})})dz=\frac{(900sin(z + 2)\cdot (3^{(\frac{1}{2})} + 9\cdot 3^{(\frac{1}{4})} - 3\cdot 3^{(\frac{3}{4})} - 3))}{(7ln(3)^{2})}|_{10}^{11}=380.99\end{align*}\)