\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{19.2}(-3cos(6ln(3x)))dx\\&\text{So the interval is }[2,19.2]\text{ the subintervals have length }\frac{19.2-(2)}{4}=\frac{43}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[2,\frac{63}{10},\frac{53}{5},\frac{149}{10}]\\&\int_{2}^{19.2}(-3cos(6ln(3x)))dx=\frac{43}{10}[(-3cos(6ln(6)))+(-3cos(6ln(\frac{189}{10})))+(-3cos(6ln(\frac{159}{5})))+(-3cos(6ln(\frac{447}{10})))]\\&\int_{2}^{19.2}(-3cos(6ln(3x)))dx=11.80\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{10.7}(-275tan(4x)^{2})dx\\&\text{So the interval is }[-1,10.7]\text{ the subintervals have length }\frac{10.7-(-1)}{3}=\frac{39}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-1,\frac{29}{10},\frac{34}{5}]\\&\int_{-1}^{10.7}(-275tan(4x)^{2})dx=\frac{39}{10}[(-275tan(4)^{2})+(-275tan(\frac{58}{5})^{2})+(-275tan(\frac{136}{5})^{2})]\\&\int_{-1}^{10.7}(-275tan(4x)^{2})dx=-7340.85\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{6.4}(14tan(14cos(6x)))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5}]\\&\int_{1}^{6.4}(14tan(14cos(6x)))dx=\frac{9}{5}[(14tan(14cos(6)))+(14tan(14cos(\frac{84}{5})))+(14tan(14cos(\frac{138}{5})))]\\&\int_{1}^{6.4}(14tan(14cos(6x)))dx=-394.53\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{22.6}(10tan(8e^{(4x)}))dx\\&\text{So the interval is }[3,22.6]\text{ the subintervals have length }\frac{22.6-(3)}{4}=\frac{49}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{79}{10},\frac{64}{5},\frac{177}{10}]\\&\int_{3}^{22.6}(10tan(8e^{(4x)}))dx=\frac{49}{10}[(10tan(8e^{(12)}))+(10tan(8e^{(\frac{158}{5})}))+(10tan(8e^{(\frac{256}{5})}))+(10tan(8e^{(\frac{354}{5})}))]\\&\int_{3}^{22.6}(10tan(8e^{(4x)}))dx=-168.87\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{6.5}(6tan(19cos(5x)))dx\\&\text{So the interval is }[5,6.5]\text{ the subintervals have length }\frac{6.5-(5)}{3}=\frac{1}{2}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,\frac{11}{2},6]\\&\int_{5}^{6.5}(6tan(19cos(5x)))dx=\frac{1}{2}[(6tan(19cos(25)))+(6tan(19cos(\frac{55}{2})))+(6tan(19cos(30)))]\\&\int_{5}^{6.5}(6tan(19cos(5x)))dx=-5.54\end{align*}\)

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval  \(\displaystyle [0,4]\) divided into \(\displaystyle 4\) sub-intervals, we'll be using rectangles with vertices at \(\displaystyle x=0,1,2,3,4\).

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is \(\displaystyle b=1\) because the rectangles are spaced \(\displaystyle 1\) unit apart. Since we're looking for the Right Riemann Sum of \(\displaystyle y=\frac{5}{x}\), we want to find the heights  of each rectangle by taking the values of each rightmost function value on each sub-interval, as follows:

\(\displaystyle f(4)=\frac{5}{4}\)

\(\displaystyle f(3)=\frac{5}{3}\)

\(\displaystyle f(2)=\frac{5}{2}\)

\(\displaystyle f(1)=\frac{5}{1}=5\)

Putting it all together, the Right Riemann Sum is

\(\displaystyle A=bh=1(\frac{5}{4}+\frac{5}{3}+\frac{5}{2}+5)=\frac{15}{12}+\frac{20}{12}+\frac{30}{12}+\frac{60}{12}=\frac{125}{12}\).

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx\\&\text{So the interval is }[1,15.8]\text{ the subintervals have length }\frac{15.8-(1)}{4}=\frac{37}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{47}{10},\frac{42}{5},\frac{121}{10},\frac{79}{5}]\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx=\frac{37}{10}[(-15sin(14e^{(4(\frac{47}{10}))}))+(-15sin(14e^{(4(\frac{42}{5}))}))+(-15sin(14e^{(4(\frac{121}{10}))}))+(-15sin(14e^{(4(\frac{79}{5}))}))]\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx=20.79\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{14}(-8sin(12e^{(x)}))dx\\&\text{So the interval is }[4,14]\text{ the subintervals have length }\frac{14-(4)}{4}=\frac{5}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{13}{2},9,\frac{23}{2},14]\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx=\frac{5}{2}[(-8sin(12e^{((\frac{13}{2}))}))+(-8sin(12e^{((9))}))+(-8sin(12e^{((\frac{23}{2}))}))+(-8sin(12e^{((14))}))]\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx=-17.23\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{6}(5x - 1000sin(3x)^{2})dx\\&\text{So the interval is }[0,6]\text{ the subintervals have length }\frac{6-(0)}{4}=\frac{3}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{3}{2},3,\frac{9}{2},6]\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx=\frac{3}{2}[(5(\frac{3}{2}) - 1000sin(3(\frac{3}{2}))^{2})+(5(3) - 1000sin(3(3))^{2})+(5(\frac{9}{2}) - 1000sin(3(\frac{9}{2}))^{2})+(5(6) - 1000sin(3(6))^{2})]\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx=-3390.69\end{align*}\)

\(\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\\&\text{So the interval is }[-4,-1.6]\text{ the subintervals have length }\frac{-1.6-(-4)}{3}=\frac{4}{5}\end{align*}\)

\(\displaystyle \begin{align*}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{16}{5},-\frac{12}{5},-\frac{8}{5}]\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\approx\frac{4}{5}[(15cos(12sin(6(-\frac{16}{5}))) + 5(-\frac{16}{5})^{3})+(15cos(12sin(6(-\frac{12}{5}))) + 5(-\frac{12}{5})^{3})+(15cos(12sin(6(-\frac{8}{5}))) + 5(-\frac{8}{5})^{3})]\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\approx-208.73\end{align*}\)