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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #1091 : Ap Calculus Ab

$\begin{align*}&f(4)=-2\\&f(13)=-11\\&f(22)=11\\&f(31)=11\\&f(40)=10\\&\text{Approximate, using a trapezoidal sum, }\int_{4}^{40}f(x)\end{align*}$

Possible Answers:

135

1080

270

Correct answer:

135

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(4)=-2\\&f(13)=-11\\&f(22)=11\\&f(31)=11\\&f(40)=10\\&\text{The total interval has a length of }36\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{4}^{40}f(x)\approx\frac{36}{2\cdot4}[(-2)+2(-11)+2(11)+2(11)+(10)]\\&\int_{4}^{40}f(x)\approx135\end{align*}$

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Example Question #1092 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-3,18]:\\&f(-3)=-12\\&f(4)=-16\\&f(11)=13\\&f(18)=-13\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

-108.5

Correct answer:

-108.5

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-3)=-12\\&f(4)=-16\\&f(11)=13\\&f(18)=-13\\&\text{The total interval has a length of }21\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-3}^{18}f(x)\approx\frac{21}{2\cdot3}[(-12)+2(-16)+2(13)+(-13)]\\&\int_{-3}^{18}f(x)\approx-108.5\end{align*}$

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Example Question #1093 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-2,34]:\\&f(-2)=8\\&f(7)=-14\\&f(16)=19\\&f(25)=2\\&f(34)=7\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

1044

130.5

261

Correct answer:

130.5

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-2)=8\\&f(7)=-14\\&f(16)=19\\&f(25)=2\\&f(34)=7\\&\text{The total interval has a length of }36\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-2}^{34}f(x)\approx\frac{36}{2\cdot4}[(8)+2(-14)+2(19)+2(2)+(7)]\\&\int_{-2}^{34}f(x)\approx130.5\end{align*}$

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Example Question #61 : Numerical Approximations To Definite Integrals

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-9,71]:\\&f(-9)=1\\&f(7)=-7\\&f(23)=2\\&f(39)=-4\\&f(55)=-3\\&f(71)=-13\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-9)=1\\&f(7)=-7\\&f(23)=2\\&f(39)=-4\\&f(55)=-3\\&f(71)=-13\\&\text{The total interval has a length of }80\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{-9}^{71}f(x)\approx\frac{80}{2\cdot5}[(1)+2(-7)+2(2)+2(-4)+2(-3)+(-13)]\\&\int_{-9}^{71}f(x)\approx-288\end{align*}$

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Example Question #1094 : Ap Calculus Ab

$\begin{align*}&\text{Approximate }\int_{0}^{54}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(0)=18\\&f(18)=-14\\&f(36)=17\\&f(54)=12\end{align*}$

Possible Answers:

1944

648

324

Correct answer:

324

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(0)=18\\&f(18)=-14\\&f(36)=17\\&f(54)=12\\&\text{The total interval has a length of }54\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{0}^{54}f(x)\approx\frac{54}{2\cdot3}[(18)+2(-14)+2(17)+(12)]\\&\int_{0}^{54}f(x)\approx324\end{align*}$

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Example Question #64 : Numerical Approximations To Definite Integrals

$\begin{align*}&\text{Approximate }\int_{2}^{34}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(2)=-11\\&f(10)=-18\\&f(18)=11\\&f(26)=7\\&f(34)=9\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(2)=-11\\&f(10)=-18\\&f(18)=11\\&f(26)=7\\&f(34)=9\\&\text{The total interval has a length of }32\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{2}^{34}f(x)\approx\frac{32}{2\cdot4}[(-11)+2(-18)+2(11)+2(7)+(9)]\\&\int_{2}^{34}f(x)\approx-8\end{align*}$

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Example Question #61 : Numerical Approximations To Definite Integrals

$\begin{align*}&f(3)=-7\\&f(13)=13\\&f(23)=12\\&f(33)=14\\&f(43)=6\\&\text{Approximate, using a trapezoidal sum, }\int_{3}^{43}f(x)\end{align*}$

Possible Answers:

385

770

3080

Correct answer:

385

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(3)=-7\\&f(13)=13\\&f(23)=12\\&f(33)=14\\&f(43)=6\\&\text{The total interval has a length of }40\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Applying the rules above, we can write:}\\&\int_{3}^{43}f(x)\approx\frac{40}{2\cdot4}[(-7)+2(13)+2(12)+2(14)+(6)]\\&\int_{3}^{43}f(x)\approx385\end{align*}$

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Example Question #62 : Numerical Approximations To Definite Integrals

$\begin{align*}&f(2)=20\\&f(6)=-19\\&f(12)=1\\&f(16)=-17\\&\text{Approximate, using a trapezoidal sum, }\int_{2}^{16}f(x)\end{align*}$

Possible Answers:

-16.5

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(2)=20\\&f(6)=-19\\&f(12)=1\\&f(16)=-17\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }4,6,4\\&\text{This means we should calculate each trapezoid separately, like in the middle summation term:}\\&\int_{2}^{16}f(x)\approx4\frac{20+(-19)}{2}+6\frac{-19+(1)}{2}+4\frac{1+(-17)}{2}\\&\int_{2}^{16}f(x)\approx-84\end{align*}$

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Example Question #67 : Numerical Approximations To Definite Integrals

$\begin{align*}&f(-4)=20\\&f(4)=18\\&f(14)=7\\&f(18)=20\\&\text{Approximate, using a trapezoidal sum, }\int_{-4}^{28}f(x)\end{align*}$

Possible Answers:

331

63.1579

2021.0526

480

Correct answer:

331

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-4)=20\\&f(4)=18\\&f(14)=7\\&f(18)=20\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }8,10,4\\&\text{This means we should calculate each trapezoid separately, like in the middle summation term:}\\&\int_{-4}^{28}f(x)\approx(8)\frac{20+(18)}{2}+(10)\frac{18+(7)}{2}+(4)\frac{7+(20)}{2}\\&\int_{-4}^{28}f(x)\approx331\end{align*}$

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Example Question #1095 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[5,29]:\\&f(5)=-6\\&f(7)=16\\&f(15)=-2\\&f(25)=-4\\&f(29)=-12\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(5)=-6\\&f(7)=16\\&f(15)=-2\\&f(25)=-4\\&f(29)=-12\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }2,8,10,4\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{5}^{29}f(x)\approx(2)\frac{-6+(16)}{2}+(8)\frac{16+(-2)}{2}+(10)\frac{-2+(-4)}{2}+(4)\frac{-4+(-12)}{2}\\&\int_{5}^{29}f(x)\approx4\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1091 : Ap Calculus Ab

Example Question #1092 : Ap Calculus Ab

Example Question #1093 : Ap Calculus Ab

Example Question #61 : Numerical Approximations To Definite Integrals

Example Question #1094 : Ap Calculus Ab

Example Question #64 : Numerical Approximations To Definite Integrals

Example Question #61 : Numerical Approximations To Definite Integrals

Example Question #62 : Numerical Approximations To Definite Integrals

Example Question #67 : Numerical Approximations To Definite Integrals

Example Question #1095 : Ap Calculus Ab

All AP Calculus AB Resources

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