Riemann Sum: Right Evaluation

AP Calculus BC · Learn by Concept

Help Questions

AP Calculus BC › Riemann Sum: Right Evaluation

1 - 10

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

CORRECT

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\&\text{So the interval is }[-1,15]\text{ the subintervals have length }\frac{15-(-1)}{4}=4\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[3,7,11,15]\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx4[(3(3) + 5sin(2e^{(2(3))}))+(3(7) + 5sin(2e^{(2(7))}))+(3(11) + 5sin(2e^{(2(11))}))+(3(15) + 5sin(2e^{(2(15))}))]\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx431.64\end{align*}$

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

CORRECT

Explanation

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

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

CORRECT

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\&\text{So the interval is }[3,4.8]\text{ the subintervals have length }\frac{4.8-(3)}{3}=\frac{3}{5}\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[\frac{18}{5},\frac{21}{5},\frac{24}{5}]\&\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\approx\frac{3}{5}[(4(\frac{18}{5}) + 16cos(13e^{(3(\frac{18}{5}))}))+(4(\frac{21}{5}) + 16cos(13e^{(3(\frac{21}{5}))}))+(4(\frac{24}{5}) + 16cos(13e^{(3(\frac{24}{5}))}))]\&\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\approx34.93\end{align*}$

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

CORRECT

Explanation

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

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

CORRECT

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\&\text{So the interval is }[-4,7.1]\text{ the subintervals have length }\frac{7.1-(-4)}{3}=\frac{37}{10}\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[-\frac{3}{10},\frac{17}{5},\frac{71}{10}]\&\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\approx\frac{37}{10}[(4(-\frac{3}{10}) - 15sin(13(-\frac{3}{10})^{2}))+(4(\frac{17}{5}) - 15sin(13(\frac{17}{5})^{2}))+(4(\frac{71}{10}) - 15sin(13(\frac{71}{10})^{2}))]\&\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\approx74.37\end{align*}$

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

CORRECT

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{-2}^{8.2}(3x - sin(15e^{(4x)}))dx\&\text{So the interval is }[-2,8.2]\text{ the subintervals have length }\frac{8.2-(-2)}{3}=\frac{17}{5}\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[\frac{7}{5},\frac{24}{5},\frac{41}{5}]\&\int_{-2}^{8.2}(3x - sin(15e^{(4x)}))dx\approx\frac{17}{5}[(3(\frac{7}{5}) - sin(15e^{(4(\frac{7}{5}))}))+(3(\frac{24}{5}) - sin(15e^{(4(\frac{24}{5}))}))+(3(\frac{41}{5}) - sin(15e^{(4(\frac{41}{5}))}))]\&\int_{-2}^{8.2}(3x - sin(15e^{(4x)}))dx\approx148.47\end{align*}$

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

CORRECT

Explanation

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

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

CORRECT

Explanation

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

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

$\frac{125}{12}$

CORRECT

$\frac{126}{12}$

$\frac{124}{12}$

$\frac{123}{12}$

$\frac{122}{12}$

Explanation

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

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

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

CORRECT

Explanation

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\&\text{with n subintervals follows the form:}\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\&\text{It is essentially a sum of n rectangles, each with a base of}\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\&\text{We're asked to approximate}\int_{2}^{4}(2x - 10tan(8tan(3x)))dx\&\text{So the interval is }[2,4]\text{ the subintervals have length }\frac{4-(2)}{4}=\frac{1}{2}\&\text{and since we are using the *right*point of each interval, the x-values are:}\&[\frac{5}{2},3,\frac{7}{2},4]\&\int_{2}^{4}(2x - 10tan(8tan(3x)))dx\approx\frac{1}{2}[(2(\frac{5}{2}) - 10tan(8tan(3(\frac{5}{2}))))+(2(3) - 10tan(8tan(3(3))))+(2(\frac{7}{2}) - 10tan(8tan(3(\frac{7}{2}))))+(2(4) - 10tan(8tan(3(4))))]\&\int_{2}^{4}(2x - 10tan(8tan(3x)))dx\approx11.06\end{align*}$

Return to subject

AI Study CoachYour personal study buddy

Instant TutoringConnect with a tutor now

Request a TutorGet matched with an expert

Live ClassesJoin interactive classes

Question of the DayDaily practice to build your skills

FlashcardsStudy and memorize key concepts

WorksheetsBuild custom practice quizzes

AI SolverStep-by-step problem solutions

Learn by ConceptMaster concepts step by step

Diagnostic TestsAssess your knowledge level

Practice TestsTest your skills with exam questions

GamesLearn through free games