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AP Calculus AB : Numerical approximations to definite integrals

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Example Questions

Example Question #906 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-8,2]:\\&f(-8)=2\\&f(-6)=-1\\&f(-4)=16\\&f(-2)=12\\&f(0)=10\\&f(2)=-18\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

210

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=2\\&f(-6)=-1\\&f(-4)=16\\&f(-2)=12\\&f(0)=10\\&f(2)=-18\\&\text{Although the total interval has a length of }10\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-8}^{2}f(x)\approx2[(2)+(-1)+(16)+(12)+(10)]\\&\int_{-8}^{2}f(x)\approx78\end{align*}$

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Example Question #907 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[5,29]:\\&f(5)=8\\&f(13)=-15\\&f(21)=9\\&f(29)=-16\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(5)=8\\&f(13)=-15\\&f(21)=9\\&f(29)=-16\\&\text{Although the total interval has a length of }24\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }8\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{5}^{29}f(x)\approx8[(8)+(-15)+(9)]\\&\int_{5}^{29}f(x)\approx16\end{align*}$

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Example Question #201 : Integrals

$\begin{align*}&f(0)=10\\&f(4)=3\\&f(8)=10\\&f(12)=-11\\&f(16)=10\\&\text{Approximate, using a left Riemann sum, }\int_{0}^{16}f(x)\end{align*}$

Possible Answers:

352

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(0)=10\\&f(4)=3\\&f(8)=10\\&f(12)=-11\\&f(16)=10\\&\text{Although the total interval has a length of }16\\&\text{Notice the points are equidistant, with }4\text{ intervals between.}\\&\text{Each equidistant interval has length: }4\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{0}^{16}f(x)\approx4[(10)+(3)+(10)+(-11)]\\&\int_{0}^{16}f(x)\approx48\end{align*}$

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Example Question #901 : Ap Calculus Ab

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-5,5]:\\&f(-5)=-5\\&f(-3)=8\\&f(-1)=4\\&f(1)=12\\&f(3)=-5\\&f(5)=-12\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-5)=-5\\&f(-3)=8\\&f(-1)=4\\&f(1)=12\\&f(3)=-5\\&f(5)=-12\\&\text{Although the total interval has a length of }10\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }2\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-5}^{5}f(x)\approx2[(-5)+(8)+(4)+(12)+(-5)]\\&\int_{-5}^{5}f(x)\approx28\end{align*}$

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Example Question #21 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{The function }f(x)\text{ has the following values on the interval }x=[-4,11]:\\&f(-4)=2\\&f(-1)=-11\\&f(2)=6\\&f(5)=-1\\&f(8)=-14\\&f(11)=12\\&\text{Approximate the integral of }f(x)\text{ on this interval using left Riemann sums.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-4)=2\\&f(-1)=-11\\&f(2)=6\\&f(5)=-1\\&f(8)=-14\\&f(11)=12\\&\text{Although the total interval has a length of }15\\&\text{Notice the points are equidistant, with }5\text{ intervals between.}\\&\text{Each equidistant interval has length: }3\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-4}^{11}f(x)\approx3[(2)+(-11)+(6)+(-1)+(-14)]\\&\int_{-4}^{11}f(x)\approx-54\end{align*}$

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Example Question #22 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&f(-2)=-17\\&f(1)=-4\\&f(4)=-16\\&f(7)=-16\\&\text{Approximate, using a left Riemann sum, }\int_{-2}^{7}f(x)\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-2)=-17\\&f(1)=-4\\&f(4)=-16\\&f(7)=-16\\&\text{Although the total interval has a length of }9\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }3\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-2}^{7}f(x)\approx3[(-17)+(-4)+(-16)]\\&\int_{-2}^{7}f(x)\approx-111\end{align*}$

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Example Question #23 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{5}^{23}f(x)\\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\\&f(5)=-3\\&f(11)=8\\&f(17)=11\\&f(23)=-3\end{align*}$

Possible Answers:

234

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(5)=-3\\&f(11)=8\\&f(17)=11\\&f(23)=-3\\&\text{Although the total interval has a length of }18\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }6\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{5}^{23}f(x)\approx6[(-3)+(8)+(11)]\\&\int_{5}^{23}f(x)\approx96\end{align*}$

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Example Question #24 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

$\begin{align*}&\text{Approximate }\int_{-8}^{19}f(x)\\&\text{Using a left Riemann sum, if }f(x)\text{ has the values:}\\&f(-8)=-10\\&f(1)=12\\&f(10)=-1\\&f(19)=11\end{align*}$

Possible Answers:

324

108

Correct answer:

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{In our case, we have function values over an interval:}\\&f(-8)=-10\\&f(1)=12\\&f(10)=-1\\&f(19)=11\\&\text{Although the total interval has a length of }27\\&\text{Notice the points are equidistant, with }3\text{ intervals between.}\\&\text{Each equidistant interval has length: }9\\&\text{Since we are using the left point of each interval, we disregard the larst function value:}\\&\int_{-8}^{19}f(x)\approx9[(-10)+(12)+(-1)]\\&\int_{-8}^{19}f(x)\approx9\end{align*}$

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Example Question #21 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Let $f(x) = \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x$ .

A relative maximum of the graph of can be located at:

Possible Answers:

The graph of has no relative maximum.

$x=- \sqrt[4]{5}$

$x= \sqrt[4]{5}$

$x =- \sqrt {6}$

$x = \sqrt {6}$

Correct answer:

$x =- \sqrt {6}$

Explanation:

At a relative minimum (c,f(c)) of the graph f(x) , it will hold that f'(c) = 0 and f''(c) < 0 .

First, find f'(x) . Using the sum rule,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x \right )$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{6}{5} x^{5} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{5}{3}x^{3} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( 30x \right )$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$f'(x) = \frac{1}{7} \cdot 7 x^{6} - \frac{6}{5} \cdot 5 x^{4} + \frac{5}{3} \cdot 3 x^{2} - 30x$

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

Set this equal to 0:

$x^{6} - 6 x^{4} +5 x^{2} - 30 = 0$

$(x^{6} - 6 x^{4} )+(5 x^{2} - 30) = 0$

$x^{4} (x^{2} - 6 )+ 5 (x^{2} - 6 ) = 0$

$(x^{4}+ 5 ) (x^{2} - 6 ) = 0$

Either:

$x^{4}+5 = 0$ , in which case, $x^{4} = -5$ ; this equation has no real solutions.

$x^{2}- 6=0$ has two real solutions, $-\sqrt {6}$ and $\sqrt {6}$ .

Now take the second derivative, again using the sum rule:

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} - 6 x^{4} +5 x^{2} - 30)$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} )- \frac{\mathrm{d} }{\mathrm{d} x} ( 6 x^{4} )+\frac{\mathrm{d} }{\mathrm{d} x}( 5 x^{2} )- \frac{\mathrm{d} }{\mathrm{d} x}( 30)$

Differentiate the individual terms using the Constant Multiple and Power Rules:

$f''(x) =6 x^{5} - 6 ( 4 x^{3} )+ 5( 2x )- 0$

$f''(x) =6 x^{5} -2 4 x^{3} +10x$

Substitute $\sqrt {6}$ for :

$f''( \sqrt {6}) =6 ( \sqrt {6} )^{5} -2 4 ( \sqrt {6})^{3} +10( \sqrt {6})$

$f''(-\sqrt {6}) = 216\sqrt {6}-144 \sqrt {6} +10\sqrt {6}$

$f''(-\sqrt {6}) =82\sqrt {6} >0$

Therefore, has a relative minimum at $x = \sqrt {6}$ .

Now. substitute $-\sqrt {6}$ for :

$f''(-\sqrt {6}) =6 (-\sqrt {6} )^{5} -2 4 (-\sqrt {6})^{3} +10(-\sqrt {6})$

$f''(-\sqrt {6}) =- 216\sqrt {6}+144 \sqrt {6} -10\sqrt {6}$

$f''(-\sqrt {6}) =-82\sqrt {6} < 0$

Therefore, has a relative maximum at $x =- \sqrt {6}$ .

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Example Question #911 : Ap Calculus Ab

Estimate the integral of f(x)=3x^2+1 from 0 to 3 using left-Riemann sum and 6 rectangles. Use $\Delta x=0.5$

Possible Answers:

74.25

23.125

46.25

23.625

37.125

Correct answer:

23.625

Explanation:

Because our $\Delta x$ is constant, the left Riemann sum will be

$R_{L}=\Delta x(f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5))$

$R_{L}=0.5\cdot (1+1.75+4+7.75+13+19.75)=23.625$

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AP Calculus AB : Numerical approximations to definite integrals

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #906 : Ap Calculus Ab

Example Question #907 : Ap Calculus Ab

Example Question #201 : Integrals

Example Question #901 : Ap Calculus Ab

Example Question #21 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #22 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #23 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #24 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #21 : Riemann Sums (Left, Right, And Midpoint Evaluation Points)

Example Question #911 : Ap Calculus Ab

All AP Calculus AB Resources

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