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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Riemann Sums

Given a function $y=x^{2}$ , find the Left Riemann Sum of the function on the interval [0,6] divided into three sub-intervals.

Possible Answers:

Correct answer:

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 [0,6] divided into sub-intervals, we'll be using rectangles with vertices at 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 b=2 because the rectangles are spaced units apart. Since we're looking for the Left Riemann Sum, we want to find the heights of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

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

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

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

Putting it all together, the Left Riemann Sum is

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

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Example Question #1 : Riemann Sums

Given a function $y=\frac{1}{4}x+2$ , find the Left Riemann Sum of the function on the interval [0,4] divided into four sub-intervals.

Possible Answers:

$\frac{19}{2}$

$\frac{17}{2}$

$\frac{21}{2}$

$\frac{25}{2}$

$\frac{23}{2}$

Correct answer:

$\frac{19}{2}$

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 [0,4] divided into sub-intervals, we'll be using rectangles with vertices at 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 b=1 because the rectangles are spaced unit apart. Since we're looking for the Left Riemann Sum, we want to find the heights of each rectangle by taking the values of each leftmost function value on each sub-interval, as follows:

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

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

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

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

Putting it all together, the Left Riemann Sum is

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

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Example Question #1 : Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{3}^{3.9}(-9tan(10sin(5x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

94.16

9.91

59.47

3.81

Correct answer:

9.91

Explanation:

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

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Example Question #1 : Numerical Approximations To Definite Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-2}^{11.6}(\frac{18}{3^{(4tan(4x))}})dx\\&\text{Using left points over }4\text{ intervals.}\end{align*}$

Possible Answers:

343.87

815.11

17826.44

2200.80

Correct answer:

2200.80

Explanation:

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

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Example Question #2 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{-4}^{4.4}(6cos(11e^{(2x)}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

14.43

6.25

41.86

4.19

Correct answer:

41.86

Explanation:

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

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Example Question #1 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{0}^{12}(11sin(20sin(x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

1.20

11.66

68.77

4.86

Correct answer:

11.66

Explanation:

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

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Example Question #1 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{0}^{3}(-16sin(2x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-249.10

-10.85

-30.38

-151.89

Correct answer:

-30.38

Explanation:

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

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Example Question #111 : Ap Calculus Bc

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{4}^{12.7}(-17cos(20x^{2}))dx\end{align*}$

Possible Answers:

4.20

91.80

1.91

16.39

Correct answer:

16.39

Explanation:

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

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Example Question #1 : Riemann Sum: Left Evaluation

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{0}^{9}(-\frac{4}{4^e^{(6x)}})dx\end{align*}$

Possible Answers:

-10.80

-0.31

-0.44

-3.00

Correct answer:

-3.00

Explanation:

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

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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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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #2 : Riemann Sums

Example Question #1 : Riemann Sums

Example Question #1 : Integrals

Example Question #1 : Numerical Approximations To Definite Integrals

Example Question #2 : Riemann Sum: Left Evaluation

Example Question #1 : Riemann Sum: Left Evaluation

Example Question #1 : Riemann Sum: Left Evaluation

Example Question #111 : Ap Calculus Bc

Example Question #1 : Riemann Sum: Left Evaluation

Example Question #11 : Riemann Sum: Left Evaluation

All AP Calculus BC Resources

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