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AP Calculus AB : Numerical approximations to definite integrals

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

Use Left Riemann sums with 4 subintervals to approximate the area between the x-axis, f(x)=x^4 +1 , x=-2 , and x=2 .

Possible Answers:

Correct answer:

Explanation:

To use left Riemann sums, we need to use the following formula:

$Area \approx \sum_{i=1}^n f(x_i)\Delta x$ .

where is the number of subintervals, (4 in our problem),

is the "counter" that denotes which subinterval we are working with,(4 subintervals mean that will be 1, 2, 3, and then 4)

f(x_i) is the function value when you plug in the "i-th" x value, (i-th in this case will be 1-st, 2-nd, 3-rd, and 4-th)

, $\Delta x$ is the width of each subinterval, which we will determine shortly.

and $\sum_{i=1}^n$ means add all versions together (for us that means add up 4 versions).

This fancy equation approximates using boxes. We can rewrite this fancy equation by writing $f(x_i)\Delta x$ , 4 times; 1 time each for i=1 , i=2 , i=3 , and i=4 . This gives us

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

Think of $\Delta x$ as the base of each box, and f(x_1) as the height of the 1st box.

This is basically Area = (base)(height) , 4 times, and then added together.

Now we need to determine what $\Delta x, x_1, x_2, x_3,$ and x_4 are.

To find $\Delta x$ we find the total length between the beginning and ending x values, which are given in the problem as x=-2 and x=2 . We then split this total length into 4 pieces, since we are told to use 4 subintervals.

In short, $\Delta x = \frac{b - a}{n}= \frac{(2)- (-2)}{4} = 1$

Now we need to find the x values that are the left endpoints of each of the 4 subintervals. Left endpoints because we are doing Left Riemann sums.

The left most x value happens to be the smaller of the overall endpoints given in the question. In other words, since we only care about the area from x=-2 to x=2 , we'll just use the smaller one, , for our first x_i .

Now we know that x_1 = -2 .

To find the next endpoint, x_2 , just increase the first x by the length of the subinterval, which is $\Delta x = 1$ . In other words

$x_2 = x_1 + \Delta x$

x_2 = -2 + 1

x_2 = -1

Add the $\Delta x$ again to get x_3

$x_3 = x_2 + \Delta x$

x_3 = -1 + 1

x_3 = 0

And repeat to find x_4

$x_4 = x_3 + \Delta x$

x_4 = 0 + 1

x_4 = 1

Now that we have all the pieces, we can plug them in.

$f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x + f(x_4)\Delta x$

$f(-2)\cdot1 + f(-1) \cdot (1)+ f(0)\cdot (1) + f(1)\cdot (1)$

plug each value into f(x)=x^4 +1 and then simplify.

[(-2)^4+1](1)+[(-1)^4+1](1)+ [(0)^4+1](1)+[(1)^4+1](1)

[16 + 1]+ [1 + 1]+ [0+ 1]+ [1 + 1]

17 + 2 + 1 + 2

This is the final answer, which is an approximation of the area under the function.

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Example Question #1 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{Using }2\text{ intervals.}\end{align*}$

Possible Answers:

-3.29

-63.97

-10.84

-105.17

Correct answer:

-10.84

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{So the interval is }[-4,1.4]\text{ the subintervals have length }\frac{1.4-(-4)}{2}=\frac{27}{10}\\&\text{and the x-values are:}\\&[-4,-\frac{13}{10},\frac{7}{5}]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx\frac{27}{20}[(11sin(17cos((-4))))+2(11sin(17cos((-\frac{13}{10}))))+(11sin(17cos((\frac{7}{5}))))]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx-10.84\end{align*}$

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Example Question #1 : Trapezoidal Sums

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

Possible Answers:

-1.68

-0.60

-0.90

-5.20

Correct answer:

-5.20

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{So the interval is }[0,2]\text{ the subintervals have length }\frac{2-(0)}{2}=1\\&\text{and the x-values are:}\\&[0,1,2]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx\frac{1}{2}[(6cos(20e^{((0))}))+2(6cos(20e^{((1))}))+(6cos(20e^{((2))}))]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx-5.20\end{align*}$

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Example Question #3 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\end{align*}$

Possible Answers:

114.16

35.68

6.37

4.20

Correct answer:

35.68

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\\&\text{So the interval is }[-1,8.9]\text{ the subintervals have length }\frac{8.9-(-1)}{3}=\frac{33}{10}\\&\text{and the x-values are:}\\&[-1,\frac{23}{10},\frac{28}{5},\frac{89}{10}]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx\frac{33}{20}[(5sin(9e^{((-1))}))+2(5sin(9e^{((\frac{23}{10}))}))+2(5sin(9e^{((\frac{28}{5}))}))+(5sin(9e^{((\frac{89}{10}))}))]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx35.68\end{align*}$

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Example Question #1 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-33.45

-93.67

-16.73

-138.84

Correct answer:

-16.73

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5},\frac{32}{5}]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx\frac{9}{10}[(-6cos(19(1)^{2}))+2(-6cos(19(\frac{14}{5})^{2}))+2(-6cos(19(\frac{23}{5})^{2}))+(-6cos(19(\frac{32}{5})^{2}))]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx-16.73\end{align*}$

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Example Question #5 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

3.13

7.65

27.55

5.51

Correct answer:

27.55

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{So the interval is }[2,6.5]\text{ the subintervals have length }\frac{6.5-(2)}{3}=\frac{3}{2}\\&\text{and the x-values are:}\\&[2,\frac{7}{2},5,\frac{13}{2}]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx\frac{3}{4}[(17cos(2sin(2(2))))+2(17cos(2sin(2(\frac{7}{2}))))+2(17cos(2sin(2(5))))+(17cos(2sin(2(\frac{13}{2}))))]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx27.55\end{align*}$

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Example Question #6 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{Using }2\text{ intervals.}\end{align*}$

Possible Answers:

-88.69

-2160.41

-130.70

-744.97

Correct answer:

-744.97

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{So the interval is }[0,4.4]\text{ the subintervals have length }\frac{4.4-(0)}{2}=\frac{11}{5}\\&\text{and the x-values are:}\\&[0,\frac{11}{5},\frac{22}{5}]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx\frac{11}{10}[(-196cos(3(0))^{2})+2(-196cos(3(\frac{11}{5}))^{2})+(-196cos(3(\frac{22}{5}))^{2})]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx-744.97\end{align*}$

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Example Question #7 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\end{align*}$

Possible Answers:

-0.29

-18.44

-0.56

-1.86

Correct answer:

-1.86

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\\&\text{So the interval is }[3,7.2]\text{ the subintervals have length }\frac{7.2-(3)}{3}=\frac{7}{5}\\&\text{and the x-values are:}\\&[3,\frac{22}{5},\frac{29}{5},\frac{36}{5}]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx\frac{7}{10}[(11cos(6e^{(4(3))}))+2(11cos(6e^{(4(\frac{22}{5}))}))+2(11cos(6e^{(4(\frac{29}{5}))}))+(11cos(6e^{(4(\frac{36}{5}))}))]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx-1.86\end{align*}$

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Example Question #1 : Trapezoidal Sums

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

Possible Answers:

-12.19

-3.05

-0.38

-0.59

Correct answer:

-3.05

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\\&\text{So the interval is }[-5,-3.8]\text{ the subintervals have length }\frac{-3.8-(-5)}{2}=\frac{3}{5}\\&\text{and the x-values are:}\\&[-5,-\frac{22}{5},\frac{11}{5}]\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\approx\frac{3}{10}[(-16sin(9e^{(5(-5))}))+2(-16sin(9e^{(5(-\frac{22}{5}))}))+(-16sin(9e^{(5(\frac{11}{5}))}))]\\&\int_{-5}^{-3.8}(-16sin(9e^{(5x)}))dx\approx-3.05\end{align*}$

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Example Question #1 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\end{align*}$

Possible Answers:

9698.21

53340.17

4849.11

91163.20

Correct answer:

9698.21

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{6.5}(2700sin(3x)^{2})dx\\&\text{So the interval is }[-1,6.5]\text{ the subintervals have length }\frac{6.5-(-1)}{3}=\frac{5}{2}\\&\text{and the x-values are:}\\&[-1,\frac{3}{2},4,\frac{13}{2}]\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\approx\frac{5}{4}[(2700sin(3(-1))^{2})+2(2700sin(3(\frac{3}{2}))^{2})+2(2700sin(3(4))^{2})+(2700sin(3(\frac{13}{2}))^{2})]\\&\int_{-1}^{6.5}(2700sin(3x)^{2})dx\approx9698.21\end{align*}$

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

Example Question #1 : Trapezoidal Sums

Example Question #1 : Trapezoidal Sums

Example Question #3 : Trapezoidal Sums

Example Question #1 : Trapezoidal Sums

Example Question #5 : Trapezoidal Sums

Example Question #6 : Trapezoidal Sums

Example Question #7 : Trapezoidal Sums

Example Question #1 : Trapezoidal Sums

Example Question #1 : Trapezoidal Sums

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