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AP Calculus BC : 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 #1 : Riemann Sum: Right Evaluation

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

Possible Answers:

647.06

74.37

21.25

505.75

Correct answer:

74.37

Explanation:

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

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Example Question #7 : Riemann Sum: Right Evaluation

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

Possible Answers:

431.64

3927.89

1553.89

2417.16

Correct answer:

431.64

Explanation:

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

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Example Question #8 : Riemann Sum: Right Evaluation

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

Possible Answers:

3.88

34.93

136.22

6.35

Correct answer:

34.93

Explanation:

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

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Example Question #1 : Riemann Sum: Right Evaluation

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

Possible Answers:

57.29

527.09

2635.46

1265.02

Correct answer:

527.09

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\\&\text{So the interval is }[-5,15]\text{ the subintervals have length }\frac{15-(-5)}{4}=5\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[0,5,10,15]\\&\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\approx5[(4(0) + 18sin(tan(5(0))))+(4(5) + 18sin(tan(5(5))))+(4(10) + 18sin(tan(5(10))))+(4(15) + 18sin(tan(5(15))))]\\&\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\approx527.09\end{align*}$

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Example Question #21 : Numerical Approximations To Definite Integrals

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

Possible Answers:

4.26

1.38

1.68

11.06

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 #21 : 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:

26.99

18.11

42.42

148.47

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 #11 : Riemann Sum: Right Evaluation

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

Possible Answers:

208.63

114.90

30.24

241.89

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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Example Question #21 : Integrals

Approximate

$\int_{1}^{5} \frac{ \ln x \; \mathrm {d}x }{x^2}$

using the midpoint rule with n=4 . Round your answer to three decimal places.

Possible Answers:

0.446

0.503

0.382

0.414

0.828

Correct answer:

0.503

Explanation:

The interval $\left [ 1,5 \right ]$ is 4 units in width; the interval is divided evenly into four subintervals $\Delta x = 4\div 4 = 1$ units in width, with their midpoints shown:

$\left [ 1,2 \right ] : x _{1} = 1.5$

$\left [2,3 \right ]: x _{2} = 2.5$

$\left [3,4 \right ] : x _{3} = 3.5$

$\left [ 4,5 \right ] : x _{4} = 4.5$

The midpoint rule requires us to calculate:

$M = \Delta x \left [ f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) \right ]$

where $\Delta x = 1$ and $f(x) = \frac{ \ln x }{x^2}$

Evaluate $f(x) = \frac{ \ln x }{x^2}$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }$ :

$f(1.5) = \frac{ \ln 1.5 }{1.5^2} \approx \frac{0.4055}{2.25} \approx 0.1802$

$f(2.5) = \frac{ \ln 2.5 }{2.5^2} \approx \frac{0.9163}{6.25} \approx 0.1466$

$f(3.5) = \frac{ \ln 3.5 }{3.5^2} \approx \frac{1.2528}{12.25} \approx 0.1023$

$f(4.5) = \frac{ \ln 4.5 }{4.5^2} \approx \frac{1.5041}{20.25} \approx 0.0743$

$M =1 \left (0.1802 + 0.1466 +0.1023 + 0.0743 \right )$

= 0.503

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Example Question #131 : Ap Calculus Bc

Approximate

$\int_{1}^{2} x^{2} \ln x \; \mathrm {d}x$

using the midpoint rule with n = 5 . Round your answer to three decimal places.

Possible Answers:

0.806

Cannot be determined

1.360

1.083

1.064

Correct answer:

1.064

Explanation:

The interval $\left [ 1,2 \right ]$ is 1 unit in width; the interval is divided evenly into five subintervals $\Delta x = 1 \div 5 = 0.2$ units in width, with their midpoints shown:

$\left [ 1, 1.2 \right ]; x_{1} = 1.1$

$\left [ 1.2, 1.4 \right ]; x_{2} = 1.3$

$\left [1.4, 1.6 \right ]; x_{3} =1.5$

$\left [ 1.6, 1.8 \right ]; x_{4} = 1.7$

$\left [1.8, 2 \right ];x_{5} = 1.9$

The midpoint rule requires us to calculate:

$M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\Delta x = 0.2$ and $f(x) =x^{2} \ln x$

Evaluate $f(x) =x^{2} \ln x$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

$f(1.1) = 1.1 ^{2} \cdot \ln 1.1 \approx 1.1 ^{2} \cdot 0.0953 \approx 0.1153$

$f(1.3) = 1.3^{2} \cdot \ln 1.3 \approx 1.3 ^{2} \cdot 0.2624 \approx 0.4434$

$f(1.5) = 1.5 ^{2} \cdot \ln 1.5 \approx 1.5 ^{2} \cdot 0.4055 \approx 0.9123$

$f(1.7) = 1.7 ^{2} \cdot \ln 1.7\approx 1.7 ^{2} \cdot 0.5306 \approx 1.5335$

$f(1.9) = 1.9 ^{2} \cdot \ln 1.9 \approx 1.9 ^{2} \cdot 0.6419 \approx 2.3171$

$M \approx 0.2 \cdot \left ( 0.1153 +0.4434 + 0.9123 + 1.5335 + 2.3171 \right ) \approx 1.064$

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Example Question #132 : Ap Calculus Bc

Approximate

$\int_{0}^{\frac{\pi}{2}} x^{2} \cos x \; \mathrm {d}x$

using the midpoint rule with n = 5 . Round your answer to three decimal places.

Possible Answers:

0.447

0.478

0.470

None of the other choices are correct.

0.467

Correct answer:

0.478

Explanation:

The interval $\left [ 0, \frac{\pi}{2}\right ]$ is $\frac{\pi}{2}$ units in width; the interval is divided evenly into five subintervals $\Delta x = \frac{\pi}{2} \div 5 = \frac{\pi}{10}$ units in width, with their midpoints shown:

$\left [ 0, \frac{\pi}{10}\right ]; x_{1} = \frac{\pi}{20}$

$\left [ \frac{\pi}{10},\frac{\pi}{5} \right ]; x_{2} = \frac{3 \pi}{20}$

$\left [ \frac{\pi}{5},\frac{3\pi}{10} \right ]; x_{3} =\frac{5 \pi}{20} = \frac{ \pi}{4}$

$\left [ \frac{3\pi}{10}, \frac{2 \pi}{5} \right ]; x_{4} =\frac{7 \pi}{20}$

$\left [ \frac{2 \pi}{5} , \frac{\pi}{2} \right ];x_{5} = \frac{9 \pi}{20}$

The midpoint rule requires us to calculate:

$M = \Delta x \left ( f (x_{1} )+ f (x_{2}) + f (x_{3}) + f (x_{4}) + f (x_{5}) \right )$

where $\Delta x = \frac{\pi}{10}$ and $f(x) = x^{2} \cos x$

Evaluate $f(x) = x^{2} \cos x$ for each of $x = x_ { 1 }, x_ { 2 }, x_ { 3 }, x_ { 4 }, x_ { 5 }$ :

$f\left $ \frac{\pi}{20} \right$ = \left $ \frac{\pi}{20} \right$^{2} \cos \left $ \frac{\pi}{20} \right$$

$\approx 0.1571^{2} \cdot \cos 0.1571 \approx 0.1571^{2} \cdot 0.9877 \approx 0.0244$

$f\left $ \frac{3\pi}{20} \right$ = \left $ \frac{3\pi}{20} \right$^{2} \cos \left $ \frac{3\pi}{20} \right$$

$\approx 0.4712 ^{2} \cdot \cos 0.4712 \approx 0.1571^{2} \cdot 0.8910 \approx 0.1979$

$f\left $ \frac{\pi}{4} \right$ = \left $ \frac{\pi}{4} \right$^{2} \cos \left $ \frac{\pi}{4} \right$$

$\approx 0.7854^{2} \cdot \cos 0.7854 \approx 0.7854^{2} \cdot 0.7071 \approx 0.4362$

$f\left $ \frac{7\pi}{20} \right$ = \left $ \frac{7\pi}{20} \right$^{2} \cos \left $ \frac{7\pi}{20} \right$$

$\approx 1.0996^{2} \cdot \cos 1.0996 \approx 1.0996^{2} \cdot 0.4540 \approx 0.5489$

$f\left $ \frac{9\pi}{20} \right$ = \left $ \frac{9\pi}{20} \right$^{2} \cos \left $ \frac{9\pi}{20} \right$$

$\approx 1.4137 ^{2} \cdot \cos1.4137 \approx 1.4137^{2} \cdot 0.1564 \approx 0.3126$

Since $\Delta x = \frac{\pi}{10} \approx 0.3142$ ,

we can approximate as

$M \approx 0.3142 \cdot \left ( 0.0244 + 0.1979 + 0.4362 + 0.5489 + 0.3126) \right ) \approx 0.478$ .

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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 #1 : Riemann Sum: Right Evaluation

Example Question #7 : Riemann Sum: Right Evaluation

Example Question #8 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #21 : Numerical Approximations To Definite Integrals

Example Question #21 : Integrals

Example Question #11 : Riemann Sum: Right Evaluation

Example Question #21 : Integrals

Example Question #131 : Ap Calculus Bc

Example Question #132 : Ap Calculus Bc

All AP Calculus BC Resources

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