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Calculus 1 : Calculus

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #132 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{3.5}^{4.3}(cos(x))^{cos(x)}dx$ using four midpoints.

Possible Answers:

0.513+0.735i

-0.513+0.735i

-0.513-0.735i

0.513-0.735i

Correct answer:

-0.513+0.735i

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{3.5}^{4.3}(cos(x))^{cos(x)}dx$

So the interval is [3.5,4.3] , the subintervals have length $\frac{4.3-3.5}{4}=0.2$ , and since we are using the midpoints of each interval, the x-values are [3.6,3.8,4.0,4.2]

$\int_{3.5}^{4.3}(cos(x))^{cos(x)}dx\approx (0.2)[(cos(3.6))^{cos(3.6)}+(cos(3.8))^{cos(3.8)}+(cos(4.0))^{cos(4.0)}+(cos(4.2))^{cos(4.2)}]$

$\int_{3.5}^{4.3}(cos(x))^{cos(x)}dx\approx-0.513+0.735i$

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Example Question #133 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx$ using three midpoints.

Possible Answers:

0.476

1.398

0.098

2.101

0.721

Correct answer:

0.721

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx$

So the interval is [0.2,1.4] , the subintervals have length $\frac{1.4-0.2}{3}=0.4$ , and since we are using the midpoints of each interval, the x-values are [0.4,0.8,1.2]

$\int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx\approx (0.4)[(sin(0.4))^{\frac{1}{cos(0.4)}}+(sin(0.8))^{\frac{1}{cos(0.8)}}+(sin(1.2))^{\frac{1}{cos(1.2)}}]$

$\int_{0.2}^{1.4}(sin(x))^{\frac{1}{cos(x)}}dx\approx 0.721$

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Example Question #134 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{2.6}^{3.2}2.7^{sin(x^2)}dx$ using three midpoints.

Possible Answers:

4.454

1.095

6.138

0.893

2.235

Correct answer:

1.095

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{2.6}^{3.2}2.7^{sin(x^2)}dx$

So the interval is [2.6,3.2] , the subintervals have length $\frac{3.6-2.6}{3}=0.2$ , and since we are using the midpoints of each interval, the x-values are [2.7,2.9,3.1]

$\int_{2.6}^{3.2}2.7^{sin(x^2)}dx\approx (0.2)[2.7^{sin(2.7^2)}+2.7^{sin(2.9^2)}+2.7^{sin(3.1^2)}]$

$\int_{2.6}^{3.2}2.7^{sin(x^2)}dx\approx 1.095$

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Example Question #135 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))dx$ using two midpoints.

Possible Answers:

0.647

2.898

2.296

3.121

1.961

Correct answer:

2.296

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))dx$

So the interval is [0.5,2.9] , the subintervals have length $\frac{2.9-0.5}{2}=1.2$ , and since we are using the midpoints of each interval, the x-values are [1.1,2.3]

$\int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))\approx (1.2)[(sin(1.1^{tan(1.1^{cos(1.1)})}))dx+(sin(2.3^{tan(2.3^{cos(2.3)})}))]$

$\int_{0.5}^{2.9}(sin(x^{tan(x^{cos(x)})}))dx\approx 2.296$

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Example Question #136 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1}^{100} tan(\sqrt[3]{x})dx$ using three midpoints.

Possible Answers:

52.418

77.277

93.248

64.009

104.522

Correct answer:

93.248

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{1}^{100} tan(\sqrt[3]{x})dx$

So the interval is [1,100] , the subintervals have length $\frac{100-1}{3}=33$ , and since we are using the midpoints of each interval, the x-values are [17.5,50.5,83.5]

$\int_{1}^{100} tan(\sqrt[3]{x})dx\approx (33)[tan(\sqrt[3]{17.5})+tan(\sqrt[3]{50.5})+tan(\sqrt[3]{83.5})]$

$\int_{1}^{100} tan(\sqrt[3]{x})dx\approx93.248$

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Example Question #137 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx$ using three midpoints.

Possible Answers:

0.02004

0.00012

0.00078

1.00089

0.39765

Correct answer:

0.00012

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follo $\int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx\approx 0.00012$ ws the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx$

So the interval is [0.0003,0.0006] , the subintervals have length $\frac{0.0006-0.0003}{3}=0.0001$ , and since we are using the midpoints of each interval, the x-values are [0.00035,0.00045,0.00055]

$\int_{0.0003}^{0.0006}14^{cos(\frac{1}{x})}dx\approx (0.0001)[14^{cos(\frac{1}{0.00035})}+14^{cos(\frac{1}{0.00045})}+14^{cos(\frac{1}{0.00055})}]$

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Example Question #138 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{1}^{7}tan(cos(x^2+x))dx$ using three midpoints.

Possible Answers:

10.080

2.877

16.055

14.932

7.173

Correct answer:

2.877

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{1}^{7}tan(cos(x^2+x))dx$

So the interval is [1,7] , the subintervals have length $\frac{7-1}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [2,4,6]

${\int_{1}^{7}tan(cos(x^2+x))dx\approx (2)[tan(cos(2^2+2))+tan(cos(4^2+4))+tan(cos(6^2+6))]}$

$\int_{1}^{7}tan(cos(x^2+x))dx\approx 2.877$

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Example Question #139 : Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate $\int_{3}^{15}x!dx$ using three midpoints.

Possible Answers:

159687384

87181920720

6227383800

348727682880

24909535200

Correct answer:

24909535200

Explanation:

A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length $\frac{b-a}{n}$ and variable heights f(x_i) , which depend on the function value at a given point x_i .

We're asked to approximate $\int_{3}^{15}x!dx$

So the interval is [3,15] , the subintervals have length $\frac{15-3}{3}=4$ , and since we are using the midpoints of each interval, the x-values are [5,9,13]

$\int_{3}^{15}x!dx \approx (4)[5!+9!+13!]$

$\int_{3}^{15}x!dx \approx 24909535200$

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Example Question #131 : Differential Functions

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=\sqrt[4]{cos(x^2)}$ over the interval (-0.6,0.6) using four midpoints.

Possible Answers:

1.197

0.444

0.997

1.359

0.299

Correct answer:

0.997

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

Now for functions that are difficult or impossible to integrate,a Riemann sum can be used to approximate the value. A Riemann sum integral approximation over an interval [a,b] with subintervals follows the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

It is essentially a sum of rectangles each with a base of length equal to the subinterval length $\frac{b-a}{n}$ , and variable heights f(x_i) , which depend on the function value at a given point x_i .

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=\sqrt[4]{cos(x^2)}$ over the interval (-0.6,0.6)

The subintervals have length $\frac{0.6-(-0.6)}{4}=0.3$ , and since we are using the midpoints of each interval, the x-values are [-0.45,-0.15,0.15,0.45]

${\frac{1}{1.2}\int_{-0.6}^{0.6}\sqrt[4]{cos(x^2)}dx\approx (\frac{1}{4})[\sqrt[4]{cos((-0.45)^2)}+\sqrt[4]{cos((-0.15)^2)}+\sqrt[4]{cos(0.15^2)}+\sqrt[4]{cos(0.45^2)}]}$

$\frac{1}{1.2}\int_{-0.6}^{0.6}\sqrt[4]{cos(x^2)}dx\approx 0.997$

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Example Question #141 : How To Find Midpoint Riemann Sums

Utilize the method of midpoint Riemann sums to approximate the average of $f(x)=x^{x^2}$ over the interval (1,1.8) using four midpoints.

Possible Answers:

2.451

0.490

1.961

0.980

3.922

Correct answer:

2.451

Explanation:

To find the average of a function over a given interval of values [a,b] , the most precise method is to use an integral as follows:

$\frac{1}{b-a}\int_{a}^{b} f(x)dx$

$\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

Now note that when using the method of Riemann sums to find an average value of a function, the expression changes:

$\frac{1}{b-a}\int_a^b f(x)dx \approx (\frac{1}{b-a})\frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))$

$\frac{1}{b-a}\int_a^b f(x)dx \approx \frac{1}{n}(f(x_1)+f(x_2)+...+f(x_n))$

We're asked to approximate the average of $f(x)=x^{x^2}$ over the interval (1,1.8)

The subintervals have length $\frac{1.8-1}{4}=0.2$ , and since we are using the midpoints of each interval, the x-values are [1.1,1.3,1.5,1.7]

$\frac{1}{0.8}\int_{1}^{1.8}x^{x^2}dx\approx (\frac{1}{4})[1.1^{1.1^2}+1.3^{1.3^2}+1.5^{1.5^2}+1.7^{1.7^2}]$

$\frac{1}{0.8}\int_{1}^{1.8}x^{x^2}dx\approx 2.451$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #132 : Midpoint Riemann Sums

Example Question #133 : Midpoint Riemann Sums

Example Question #134 : Midpoint Riemann Sums

Example Question #135 : Midpoint Riemann Sums

Example Question #136 : Midpoint Riemann Sums

Example Question #137 : Midpoint Riemann Sums

Example Question #138 : Midpoint Riemann Sums

Example Question #139 : Midpoint Riemann Sums

Example Question #131 : Differential Functions

Example Question #141 : How To Find Midpoint Riemann Sums

All Calculus 1 Resources

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