\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=-\frac{(79e^{(-t)})}{9}|_{0.5}^{1.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=-\frac{(79e^{(-(1.5))})}{9}-(-\frac{(79e^{(-(0.5))})}{9})\\&\int_{0.5}^{1.5}(\frac{(79e^{(-t)})}{9})dt=3.37\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=\frac{(21\cdot 3^{(\frac{t}{3})})}{(2ln(3))}|_{0.5}^{2}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=\frac{(21\cdot 3^{(\frac{(2)}{3})})}{(2ln(3))}-(\frac{(21\cdot 3^{(\frac{(0.5)}{3})})}{(2ln(3))})\\&\int_{0.5}^{2}(\frac{(7\cdot 3^{(\frac{t}{3})})}{2})dt=8.40\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=-\frac{(17cos(t + 2))}{4}|_{5}^{6.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=-\frac{(17cos((6.5) + 2))}{4}-(-\frac{(17cos((5) + 2))}{4})\\&\int_{5}^{6.5}(\frac{(17sin(t + 2))}{4})dt=5.76\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=\frac{(19sin(3t))}{8}|_{1}^{6}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=\frac{(19sin(3(6)))}{8}-(\frac{(19sin(3(1)))}{8})\\&\int_{1}^{6}(\frac{(57cos(3t))}{8})dt=-2.12\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=\frac{(12e^{(t)})}{37}|_{0.5}^{3}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=\frac{(12e^{((3))})}{37}-(\frac{(12e^{((0.5))})}{37})\\&\int_{0.5}^{3}(\frac{(12e^{(t)})}{37})dt=5.98\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-\frac{(73cos(t + 1))}{11}|_{3}^{5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-\frac{(73cos((5) + 1))}{11}-(-\frac{(73cos((3) + 1))}{11})\\&\int_{3}^{5}(\frac{(73sin(t + 1))}{11})dt=-10.71\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=-\frac{(71cos(3t))}{21}|_{4}^{6}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=-\frac{(71cos(3(6)))}{21}-(-\frac{(71cos(3(4)))}{21})\\&\int_{4}^{6}(\frac{(71sin(3t))}{7})dt=0.62\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{3}^{7}(\frac{(37t)}{2})dt=\frac{(37t^{2})}{4}|_{3}^{7}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{7}(\frac{(37t)}{2})dt=\frac{(37(7)^{2})}{4}-(\frac{(37(3)^{2})}{4})\\&\int_{3}^{7}(\frac{(37t)}{2})dt=370.00\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{3}^{6.5}(\frac{9}{17})dt=\frac{(9t)}{17}|_{3}^{6.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{3}^{6.5}(\frac{9}{17})dt=\frac{(9(6.5))}{17}-(\frac{(9(3))}{17})\\&\int_{3}^{6.5}(\frac{9}{17})dt=1.85\end{align*}\)

\(\displaystyle \begin{align*}&\text{If the function defining a rate of change is known, the total}\\&\text{change over a given interval of time is only a matter of integrating}\\&\text{this rate function over this same time interval. Considering}\\&\text{the function given:}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=\frac{(50e^{(t)})}{7}|_{0.5}^{5.5}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=\frac{(50e^{((5.5))})}{7}-(\frac{(50e^{((0.5))})}{7})\\&\int_{0.5}^{5.5}(\frac{(50e^{(t)})}{7})dt=1736.02\end{align*}\)