Riemann Sum: Right Evaluation - AP Calculus BC

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Question

Given a function $y=\frac{5}{x}$$ , find the Right Riemann Sum of the function on the interval divided into four sub-intervals.

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Answer

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 divided into sub-intervals, we'll be using rectangles with vertices at .

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 because the rectangles are spaced unit apart. Since we're looking for the Right Riemann Sum of $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:

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

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

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

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

Putting it all together, the Right Riemann Sum is

$A=bh=1($\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}$$ .

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${1}^{15.8}$$(-15sin(14e^{(4x)}$))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$=\frac{37}{10}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{47}{10}$,$\frac{42}{5}$,$\frac{121}{10}$,$\frac{79}{5}$]\&\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)}$))dx=20.79\end{align}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${4}^{14}$$(-8sin(12e^{(x)}$))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$=\frac{5}{2}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{13}{2}$,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)}$))dx=-17.23\end{align}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${0}^{6}$(5x - $1000sin(3x)^{2}$)dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$=\frac{3}{2}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{3}{2}$,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}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${-4}^{-1.6}$(15cos(12sin(6x)) + $5x^{3}$)dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$

$\begin{align}\&$\text{and since we are using the rightpoint 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}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${0}^{13.6}$(2x + 4sin(3tan(5x)))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}^{13.6}$(2x + 4sin(3tan(5x)))dx\&$\text{So the interval is }$[0,13.6]$\text{ the subintervals have length }$$\frac{13.6-(0)}{4}$=\frac{17}{5}$\end{align}$

$\begin{align}\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{17}{5}$,$\frac{34}{5}$,$\frac{51}{5}$,$\frac{68}{5}$]\&\int_${0}^{13.6}$(2x + 4sin(3tan(5x)))dx\approx$\frac{17}{5}$[(2($\frac{17}{5}$) + 4sin(3tan(5($\frac{17}{5}$))))+(2($\frac{34}{5}$) + 4sin(3tan(5($\frac{34}{5}$))))+(2($\frac{51}{5}$) + 4sin(3tan(5($\frac{51}{5}$))))+(2($\frac{68}{5}$) + 4sin(3tan(5($\frac{68}{5}$))))]\&\int_${0}^{13.6}$(2x + 4sin(3tan(5x)))dx\approx214.27\end{align}$

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Question

$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of right point Riemann sums approximate the integral:}$\&\int_${-4}^{7.1}$(4x - $15sin(13x^{2}$))dx\end{align}$

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Answer

$\begin{align}&$\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}^{7.1}$(4x - $15sin(13x^{2}$))dx\&$\text{So the interval is }$[-4,7.1]$\text{ the subintervals have length }$$\frac{7.1-(-4)}{3}$=\frac{37}{10}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[-$\frac{3}{10}$,$\frac{17}{5}$,$\frac{71}{10}$]\&\int_${-4}^{7.1}$(4x - $15sin(13x^{2}$))dx\approx$\frac{37}{10}$[(4(-$\frac{3}{10}$) - $15sin(13(-$\frac{3}{10}$)^{2}$))+(4($\frac{17}{5}$) - $15sin(13($\frac{17}{5}$)^{2}$))+(4($\frac{71}{10}$) - $15sin(13($\frac{71}{10}$)^{2}$))]\&\int_${-4}^{7.1}$(4x - $15sin(13x^{2}$))dx\approx74.37\end{align}$

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Question

$\begin{align}&$\text{Calculate the right-side Riemann sums integral approximation of :}$\&\int_${-1}^{15}$(3x + $5sin(2e^{(2x)}$))dx\&$\text{Using points over }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$(3x + $5sin(2e^{(2x)}$))dx\&$\text{So the interval is }$[-1,15]$\text{ the subintervals have length }$$\frac{15-(-1)}{4}$=4\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[3,7,11,15]\&\int_${-1}^{15}$(3x + $5sin(2e^{(2x)}$))dx\approx4[(3(3) + $5sin(2e^{(2(3))}$))+(3(7) + $5sin(2e^{(2(7))}$))+(3(11) + $5sin(2e^{(2(11))}$))+(3(15) + $5sin(2e^{(2(15))}$))]\&\int_${-1}^{15}$(3x + $5sin(2e^{(2x)}$))dx\approx431.64\end{align}$

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Question

$\begin{align}&$\text{Calculate the right-side Riemann sums integral approximation of :}$\&\int_${3}^{4.8}$(4x + $16cos(13e^{(3x)}$))dx\&$\text{Using points over }$3$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}^{4.8}$(4x + $16cos(13e^{(3x)}$))dx\&$\text{So the interval is }$[3,4.8]$\text{ the subintervals have length }$$\frac{4.8-(3)}{3}$=\frac{3}{5}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{18}{5}$,$\frac{21}{5}$,$\frac{24}{5}$]\&\int_${3}^{4.8}$(4x + $16cos(13e^{(3x)}$))dx\approx$\frac{3}{5}$[(4($\frac{18}{5}$) + $16cos(13e^{(3($\frac{18}${5}$))}))+(4($\frac{21}{5}$) + $16cos(13e^{(3($\frac{21}${5}$))}))+(4($\frac{24}{5}$) + $16cos(13e^{(3($\frac{24}${5}$))}))]\&\int_${3}^{4.8}$(4x + $16cos(13e^{(3x)}$))dx\approx34.93\end{align}$

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Question

$\begin{align}&$\text{Calculate the right-side Riemann sums integral approximation of :}$\&\int_${-5}^{15}$(4x + 18sin(tan(5x)))dx\&$\text{Using points over }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}^{15}$(4x + 18sin(tan(5x)))dx\&$\text{So the interval is }$[-5,15]$\text{ the subintervals have length }$$\frac{15-(-5)}{4}$=5\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[0,5,10,15]\&\int_${-5}^{15}$(4x + 18sin(tan(5x)))dx\approx5[(4(0) + 18sin(tan(5(0))))+(4(5) + 18sin(tan(5(5))))+(4(10) + 18sin(tan(5(10))))+(4(15) + 18sin(tan(5(15))))]\&\int_${-5}^{15}$(4x + 18sin(tan(5x)))dx\approx527.09\end{align}$

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Question

$\begin{align}&$\text{Using }$4$\text{ intervals, and the method of right point Riemann sums approximate the integral:}$\&\int_${2}^{4}$(2x - 10tan(8tan(3x)))dx\end{align}$

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Answer

$\begin{align}&$\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}^{4}$(2x - 10tan(8tan(3x)))dx\&$\text{So the interval is }$[2,4]$\text{ the subintervals have length }$$\frac{4-(2)}{4}$=\frac{1}{2}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{5}{2}$,3,$\frac{7}{2}$,4]\&\int_${2}^{4}$(2x - 10tan(8tan(3x)))dx\approx$\frac{1}{2}$[(2($\frac{5}{2}$) - 10tan(8tan(3($\frac{5}{2}$))))+(2(3) - 10tan(8tan(3(3))))+(2($\frac{7}{2}$) - 10tan(8tan(3($\frac{7}{2}$))))+(2(4) - 10tan(8tan(3(4))))]\&\int_${2}^{4}$(2x - 10tan(8tan(3x)))dx\approx11.06\end{align}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${-2}^{8.2}$(3x - $sin(15e^{(4x)}$))dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}^{8.2}$(3x - $sin(15e^{(4x)}$))dx\&$\text{So the interval is }$[-2,8.2]$\text{ the subintervals have length }$$\frac{8.2-(-2)}{3}$=\frac{17}{5}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{7}{5}$,$\frac{24}{5}$,$\frac{41}{5}$]\&\int_${-2}^{8.2}$(3x - $sin(15e^{(4x)}$))dx\approx$\frac{17}{5}$[(3($\frac{7}{5}$) - $sin(15e^{(4($\frac{7}${5}$))}))+(3($\frac{24}{5}$) - $sin(15e^{(4($\frac{24}${5}$))}))+(3($\frac{41}{5}$) - $sin(15e^{(4($\frac{41}${5}$))}))]\&\int_${-2}^{8.2}$(3x - $sin(15e^{(4x)}$))dx\approx148.47\end{align}$

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Question

$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of right point Riemann sums approximate the integral:}$\&\int_${5}^{6.5}$(2x + 7\cdot $3^{(5sin(x))}$)dx\end{align}$

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Answer

$\begin{align}&$\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}$(2x + 7\cdot $3^{(5sin(x))}$)dx\&$\text{So the interval is }$[5,6.5]$\text{ the subintervals have length }$$\frac{6.5-(5)}{3}$=\frac{1}{2}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{11}{2}$,6,$\frac{13}{2}$]\&\int_${5}^{6.5}$(2x + 7\cdot $3^{(5sin(x))}$)dx\approx$\frac{1}{2}$[(2($\frac{11}{2}$) + 7\cdot $3^{(5sin(($\frac{11}${2}$)))})+(2(6) + 7\cdot $3^{(5sin((6)))}$)+(2($\frac{13}{2}$) + 7\cdot $3^{(5sin(($\frac{13}${2}$)))})]\&\int_${5}^{6.5}$(2x + 7\cdot $3^{(5sin(x))}$)dx\approx30.24\end{align}$

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Question

Given a function $y=\frac{5}{x}$$ , find the Right Riemann Sum of the function on the interval divided into four sub-intervals.

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Answer

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 divided into sub-intervals, we'll be using rectangles with vertices at .

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 because the rectangles are spaced unit apart. Since we're looking for the Right Riemann Sum of $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:

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

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

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

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

Putting it all together, the Right Riemann Sum is

$A=bh=1($\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}$$ .

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${1}^{15.8}$$(-15sin(14e^{(4x)}$))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$=\frac{37}{10}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{47}{10}$,$\frac{42}{5}$,$\frac{121}{10}$,$\frac{79}{5}$]\&\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)}$))dx=20.79\end{align}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${4}^{14}$$(-8sin(12e^{(x)}$))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$=\frac{5}{2}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{13}{2}$,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)}$))dx=-17.23\end{align}$

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Question

$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${0}^{6}$(5x - $1000sin(3x)^{2}$)dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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Answer

$\begin{align}&$\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}$=\frac{3}{2}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{3}{2}$,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}$

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$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${-4}^{-1.6}$(15cos(12sin(6x)) + $5x^{3}$)dx\&$\text{Using }$3$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}$

$\begin{align}\&$\text{and since we are using the rightpoint 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}$

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$\begin{align}&$\text{Using the method of right Riemann sums approximate the integral:}$\&\int_${0}^{13.6}$(2x + 4sin(3tan(5x)))dx\&$\text{Using }$4$\text{ intervals.}$\end{align}$

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$\begin{align}&$\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}^{13.6}$(2x + 4sin(3tan(5x)))dx\&$\text{So the interval is }$[0,13.6]$\text{ the subintervals have length }$$\frac{13.6-(0)}{4}$=\frac{17}{5}$\end{align}$

$\begin{align}\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[$\frac{17}{5}$,$\frac{34}{5}$,$\frac{51}{5}$,$\frac{68}{5}$]\&\int_${0}^{13.6}$(2x + 4sin(3tan(5x)))dx\approx$\frac{17}{5}$[(2($\frac{17}{5}$) + 4sin(3tan(5($\frac{17}{5}$))))+(2($\frac{34}{5}$) + 4sin(3tan(5($\frac{34}{5}$))))+(2($\frac{51}{5}$) + 4sin(3tan(5($\frac{51}{5}$))))+(2($\frac{68}{5}$) + 4sin(3tan(5($\frac{68}{5}$))))]\&\int_${0}^{13.6}$(2x + 4sin(3tan(5x)))dx\approx214.27\end{align}$

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$\begin{align}&$\text{Using }$3$\text{ intervals, and the method of right point Riemann sums approximate the integral:}$\&\int_${-4}^{7.1}$(4x - $15sin(13x^{2}$))dx\end{align}$

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$\begin{align}&$\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}^{7.1}$(4x - $15sin(13x^{2}$))dx\&$\text{So the interval is }$[-4,7.1]$\text{ the subintervals have length }$$\frac{7.1-(-4)}{3}$=\frac{37}{10}$\&$\text{and since we are using the rightpoint of each interval, the x-values are:}$\&[-$\frac{3}{10}$,$\frac{17}{5}$,$\frac{71}{10}$]\&\int_${-4}^{7.1}$(4x - $15sin(13x^{2}$))dx\approx$\frac{37}{10}$[(4(-$\frac{3}{10}$) - $15sin(13(-$\frac{3}{10}$)^{2}$))+(4($\frac{17}{5}$) - $15sin(13($\frac{17}{5}$)^{2}$))+(4($\frac{71}{10}$) - $15sin(13($\frac{71}{10}$)^{2}$))]\&\int_${-4}^{7.1}$(4x - $15sin(13x^{2}$))dx\approx74.37\end{align}$

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