The interval $\displaystyle [0, 8]$ divided into four sub-intervals gives rectangles with vertices of the bases at

$\displaystyle x=0,2,4,6,8$

For the Left Riemann sum, we need to find the rectangle heights which values come from the left-most function value of each sub-interval, or f(0), f(2), f(4), and f(6).

$\displaystyle f(0) = 0+7 = 7$

$\displaystyle f(2) = 2+7 = 9$

$\displaystyle f(4) = 4+7 = 11$

$\displaystyle f(6) = 6+7 = 13$

Because each sub-interval has a width of 2, the Left Riemann sum is

$\displaystyle A = bh = 2(7+9+11+13) = 80$

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 $\displaystyle [0,6]$ divided into $\displaystyle 3$ sub-intervals, we'll be using rectangles with vertices at $\displaystyle x=0,2,4,6$. 

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 $\displaystyle b=2$ because the rectangles are spaced $\displaystyle 2$ units apart. Since we're looking for the Left Riemann Sum, we want to find the heights $\displaystyle h$ of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$\displaystyle f(0)=(0)^{2}=0$

$\displaystyle f(2)=(2)^{2}=4$

$\displaystyle f(4)=(4)^{2}=16$

Putting it all together, the Left Riemann Sum is 

$\displaystyle A=bh=2(0+4+16)=2(20)=40$.

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 Left Riemann Sum, we want to find the heights $\displaystyle h$ of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

$\displaystyle f(0)=\frac{1}{4}(0)+2=0+2=2$

$\displaystyle f(1)=\frac{1}{4}(1)+2=\frac{1}{4}+\frac{8}{4}=\frac{9}{4}$

$\displaystyle f(2)=\frac{1}{4}(2)+2=\frac{1}{2}+\frac{4}{2}=\frac{5}{2}$

$\displaystyle f(3)=\frac{1}{4}(3)+2=\frac{3}{4}+\frac{8}{4}=\frac{11}{4}$

Putting it all together, the Left Riemann Sum is 

$\displaystyle A=bh=1(2+\frac{9}{4}+\frac{5}{2}+\frac{11}{4})=\frac{8}{4}+\frac{9}{4}+\frac{10}{4}+\frac{11}{4}=\frac{38}{4}=\frac{19}{2}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{3.9}(-9tan(10sin(5x)))dx\\&\text{So the interval is }[3,3.9]\text{ the subintervals have length }\frac{3.9-(3)}{3}=\frac{3}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{33}{10},\frac{18}{5}]\\&\int_{3}^{3.9}(-9tan(10sin(5x)))dx=\frac{3}{10}[(-9tan(10sin(15)))+(-9tan(10sin(\frac{33}{2})))+(-9tan(10sin(18)))]\\&\int_{3}^{3.9}(-9tan(10sin(5x)))dx=9.91\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\\&\text{So the interval is }[-2,11.6]\text{ the subintervals have length }\frac{11.6-(-2)}{4}=\frac{17}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-2,\frac{7}{5},\frac{24}{5},\frac{41}{5}]\\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=\frac{17}{5}[(18\cdot 3^{(4tan(8))})+(\frac{18}{3^{(4tan(\frac{28}{5}))}})+(\frac{18}{3^{(4tan(\frac{96}{5}))}})+(\frac{18}{3^{(4tan(\frac{164}{5}))}})]\\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx=2200.80\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{4.4}(6cos(11e^{(2x)}))dx\\&\text{So the interval is }[-4,4.4]\text{ the subintervals have length }\frac{4.4-(-4)}{3}=\frac{14}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-4,-\frac{6}{5},\frac{8}{5}]\\&\int_{-4}^{4.4}(6cos(11e^{(2x)}))dx=\frac{14}{5}[(6cos(11e^{(-8)}))+(6cos(11e^{(-\frac{12}{5})}))+(6cos(11e^{(\frac{16}{5})}))]\\&\int_{-4}^{4.4}(6cos(11e^{(2x)}))dx=41.86\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{12}(11sin(20sin(x)))dx\\&\text{So the interval is }[0,12]\text{ the subintervals have length }\frac{12-(0)}{3}=4\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,4,8]\\&\int_{0}^{12}(11sin(20sin(x)))dx=4[(0)+(11sin(20sin(4)))+(11sin(20sin(8)))]\\&\int_{0}^{12}(11sin(20sin(x)))dx=11.66\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{3}(-16sin(2x^{2}))dx\\&\text{So the interval is }[0,3]\text{ the subintervals have length }\frac{3-(0)}{3}=1\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,1,2]\\&\int_{0}^{3}(-16sin(2x^{2}))dx=1[(0)+(-16sin(2))+(-16sin(8))]\\&\int_{0}^{3}(-16sin(2x^{2}))dx=-30.38\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{12.7}(-17cos(20x^{2}))dx\\&\text{So the interval is }[4,12.7]\text{ the subintervals have length }\frac{12.7-(4)}{3}=\frac{29}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[4,\frac{69}{10},\frac{49}{5}]\\&\int_{4}^{12.7}(-17cos(20x^{2}))dx=\frac{29}{10}[(-17cos(320))+(-17cos(\frac{4761}{5}))+(-17cos(\frac{9604}{5}))]\\&\int_{4}^{12.7}(-17cos(20x^{2}))dx=16.39\end{align*}$

$\displaystyle \begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{9}(-\frac{4}{4^e^{(6x)}})dx\\&\text{So the interval is }[0,9]\text{ the subintervals have length }\frac{9-(0)}{3}=3\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,3,6]\\&\int_{0}^{9}(-\frac{4}{4^e^{(6x)}})dx=3[(-1)+(-\frac{4}{4^e^{(18)}})+(-\frac{4}{4^e^{(36)}})]\\&\int_{0}^{9}(-\frac{4}{4^e^{(6x)}})dx=-3.00\end{align*}$