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AP Calculus BC : Riemann Sum: Left Evaluation

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Example Questions

Example Question #11 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{2}^{19.2}(-3cos(6ln(3x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

11.80

1.46

3.19

1.81

Correct answer:

11.80

Explanation:

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

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Example Question #12 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{-1}^{10.7}(-275tan(4x)^{2})dx\end{align*}$

Possible Answers:

-22756.63

-7340.85

-60194.95

-1048.69

Correct answer:

-7340.85

Explanation:

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

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Example Question #13 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{1}^{6.4}(14tan(14cos(6x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-394.53

-146.12

-2012.10

-41.53

Correct answer:

-394.53

Explanation:

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

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Example Question #14 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{3}^{22.6}(10tan(8e^{(4x)}))dx\end{align*}$

Possible Answers:

-506.60

-25.59

-168.87

-19.64

Correct answer:

-168.87

Explanation:

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

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Example Question #15 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{6.5}(6tan(19cos(5x)))dx\end{align*}$

Possible Answers:

-0.91

-1.38

-5.54

-0.63

Correct answer:

-5.54

Explanation:

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

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AP Calculus BC : Riemann Sum: Left Evaluation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #11 : Riemann Sum: Left Evaluation

Example Question #12 : Riemann Sum: Left Evaluation

Example Question #13 : Riemann Sum: Left Evaluation

Example Question #14 : Riemann Sum: Left Evaluation

Example Question #15 : Riemann Sum: Left Evaluation

All AP Calculus BC Resources

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