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AP Calculus BC : Riemann Sum: Right Evaluation

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Example Question #11 : Riemann Sum: Right Evaluation

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

Possible Answers:

4.26

1.38

11.06

1.68

Correct answer:

11.06

Explanation:

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

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Example Question #22 : Integrals

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

Possible Answers:

18.11

148.47

26.99

42.42

Correct answer:

148.47

Explanation:

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

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Example Question #23 : Integrals

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

Possible Answers:

30.24

241.89

114.90

208.63

Correct answer:

30.24

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{6.5}(2x + 7\cdot 3^{(5sin(x))})dx\\&\text{So the interval is }[5,6.5]\text{ the subintervals have length }\frac{6.5-(5)}{3}=\frac{1}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{11}{2},6,\frac{13}{2}]\\&\int_{5}^{6.5}(2x + 7\cdot 3^{(5sin(x))})dx\approx\frac{1}{2}[(2(\frac{11}{2}) + 7\cdot 3^{(5sin((\frac{11}{2})))})+(2(6) + 7\cdot 3^{(5sin((6)))})+(2(\frac{13}{2}) + 7\cdot 3^{(5sin((\frac{13}{2})))})]\\&\int_{5}^{6.5}(2x + 7\cdot 3^{(5sin(x))})dx\approx30.24\end{align*}$

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AP Calculus BC : Riemann Sum: Right Evaluation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

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Example Question #11 : Riemann Sum: Right Evaluation

Example Question #22 : Integrals

Example Question #23 : Integrals

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