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AP Calculus AB : Integrals

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Example Questions

Example Question #291 : 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 #292 : 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:

3080

385

770

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 #293 : 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 #294 : 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:

63.1579

2021.0526

331

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 #295 : Integrals

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

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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Example Question #296 : Integrals

$\begin{align*}&\text{Approximate }\int_{2}^{27}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(2)=-8\\&f(4)=-13\\&f(14)=-7\\&f(16)=-12\\&f(22)=17\end{align*}$

Possible Answers:

-30.692

-767.2991

-171.875

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(2)=-8\\&f(4)=-13\\&f(14)=-7\\&f(16)=-12\\&f(22)=17\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }2,10,2,6\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{2}^{27}f(x)\approx(2)\frac{-8+(-13)}{2}+(10)\frac{-13+(-7)}{2}+(2)\frac{-7+(-12)}{2}+(6)\frac{-12+(17)}{2}\\&\int_{2}^{27}f(x)\approx-125\end{align*}$

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Example Question #297 : Integrals

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-4,8]:\\&f(-4)=20\\&f(0)=13\\&f(2)=-1\\&f(8)=16\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

120

123

360

Correct answer:

123

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-4)=20\\&f(0)=13\\&f(2)=-1\\&f(8)=16\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }4,2,6\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-4}^{8}f(x)\approx(4)\frac{20+(13)}{2}+(2)\frac{13+(-1)}{2}+(6)\frac{-1+(16)}{2}\\&\int_{-4}^{8}f(x)\approx123\end{align*}$

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Example Question #298 : Integrals

$\begin{align*}&f(-9)=9\\&f(1)=6\\&f(9)=14\\&f(13)=-4\\&f(23)=10\\&\text{Approximate, using a trapezoidal sum, }\int_{-9}^{23}f(x)\end{align*}$

Possible Answers:

205

816

25.5

204

Correct answer:

205

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-9)=9\\&f(1)=6\\&f(9)=14\\&f(13)=-4\\&f(23)=10\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }10,8,4,10\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-9}^{23}f(x)\approx(10)\frac{9+(6)}{2}+(8)\frac{6+(14)}{2}+(4)\frac{14+(-4)}{2}+(10)\frac{-4+(10)}{2}\\&\int_{-9}^{23}f(x)\approx205\end{align*}$

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Example Question #31 : Trapezoidal Sums

$\begin{align*}&\text{Approximate }\int_{-5}^{17}f(x)\\&\text{Using the trapezoidal rule, if }f(x)\text{ has the values:}\\&f(-5)=5\\&f(1)=-6\\&f(11)=11\\&f(17)=-4\end{align*}$

Possible Answers:

5.5

121

40.3333

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(-5)=5\\&f(1)=-6\\&f(11)=11\\&f(17)=-4\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }6,10,6\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{-5}^{17}f(x)\approx(6)\frac{5+(-6)}{2}+(10)\frac{-6+(11)}{2}+(6)\frac{11+(-4)}{2}\\&\int_{-5}^{17}f(x)\approx43\end{align*}$

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Example Question #32 : Trapezoidal Sums

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[1,32]:\\&f(1)=15\\&f(3)=-9\\&f(7)=9\\&f(13)=-15\\&f(17)=14\\&\text{Approximate the integral of }f(x)\text{ on this interval using the trapezoidal rule.}\end{align*}$

Possible Answers:

-3.875

-0.74519

-23.101

Correct answer:

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're given a set of values:}\\&f(1)=15\\&f(3)=-9\\&f(7)=9\\&f(13)=-15\\&f(17)=14\\&\text{Notice that the points are not equidistant, with varying interval lengths between}\\&\text{These lengths in sequence are: }2,4,6,4\\&\text{This means we should calculate each trapezoid separately: }\sum_{i=1}^{n}(x_{i+1}-x_i)\frac{f(x_{i+1})+f(x_i)}{2}\\&\int_{1}^{32}f(x)\approx(2)\frac{15+(-9)}{2}+(4)\frac{-9+(9)}{2}+(6)\frac{9+(-15)}{2}+(4)\frac{-15+(14)}{2}\\&\int_{1}^{32}f(x)\approx-14\end{align*}$

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AP Calculus AB : Integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #291 : Integrals

Example Question #292 : Integrals

Example Question #293 : Integrals

Example Question #294 : Integrals

Example Question #295 : Integrals

Example Question #296 : Integrals

Example Question #297 : Integrals

Example Question #298 : Integrals

Example Question #31 : Trapezoidal Sums

Example Question #32 : Trapezoidal Sums

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