We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

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

Example Questions

Example Question #92 : How To Find Midpoint Riemann Sums

Approximate the integral $\int_3^7 1.02^{x^3}$ using the method of midpoint Riemann sums and four midpoints.

Possible Answers:

11.2

23.7

8.9

265.4

39.2

Correct answer:

265.4

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 are approximating $\int_3^7 1.02^{x^3}$

So the interval is [3,7] , the subintervals have length $\frac{7-3}{4}=1$ , and since we are using the midpoints of each interval, the x-values are [3.5,4.5,5.5,6.5]

$\int_3^7 1.02^{x^3}\approx 1[1.02^{3.5^3}+1.02^{4.5^3}+1.02^{5.5^3}+1.02^{6.5^3}]$

$\int_3^7 1.02^{x^3} \approx 265.4$

Report an Error

Example Question #93 : How To Find Midpoint Riemann Sums

Using the method of midpoint Riemann sums, approximate the integral $\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}})$ using four midpoints.

Possible Answers:

0.149+0.115i

0.149-0.115i

0.115-0.149i

0.231+i

0.115+0.149i

Correct answer:

0.115+0.149i

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 approximating $\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}})$

So the interval is [-1.2,2.4] , the subintervals have length $\frac{2.4-(-1.2)}{4}=0.9$ , and since we are using the midpoints of each interval, the x-values are [-0.75,0.15,1.05,1.95]

${\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}})\approx 0.9[cos(1.7^{(-0.75)^{1.3}})+cos(1.7^{(0.15)^{1.3}})+cos(1.7^{(1.05)^{1.3}})+cos(1.7^{(1.95)^{1.3}})]}$

$\int_{-1.2}^{2.4}cos(1.7^{x^{1.3}}) \approx 0.115+0.149i$

Report an Error

Example Question #94 : Midpoint Riemann Sums

Using the method of midpoint of Reimann sums, approximate the integral $\int_0^3 tan(x^2)dx$ using three midpoints.

Possible Answers:

0.49

2.31

-0.81

0.78

-1.02

Correct answer:

-1.02

Explanation:

A Reimann 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 approximating $\int_0^3 tan(x^2)dx$

So the interval is [0,3] , the subintervals have length $\frac{3-0}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [0.5,1.5,2.5]

$\int_0^3 tan(x^2)dx \approx 1[ tan(0.5^2)+ tan(1.5^2)+ tan(2.5^2)]=-1.02$

Report an Error

Example Question #94 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_1^5 ln(tan(2^x))dx$ using four midpoints.

Possible Answers:

-0.64 +9.42i

9.42-0.64 i

-0.64 -9.42i

9.42+0.64 i

-9.42-0.64 i

Correct answer:

-0.64 +9.42i

Explanation:

A Reimann 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 approximating $\int_1^5 ln(tan(2^x))dx$

So the interval is [1,5] , the subintervals have length $\frac{5-1}{4}=1$ , and since we are using the midpoints of each interval, the x-values are [1.5,2.5,3.5,4.5]

$\int_1^5 ln(tan(2^x))dx \approx 1[ln(tan(2^{1.5}))+ln(tan(2^{2.5}))+ln(tan(2^{3.5}))+ln(tan(2^{4.5}))]$

$\int_1^5 ln(tan(2^x))dx \approx -0.64 +9.42i$

Report an Error

Example Question #91 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_2^8 \frac{1}{tan(3^x)}dx$ using two midpoints.

Possible Answers:

4.5

-19.5

18.3

6.1

-6.5

Correct answer:

-19.5

Explanation:

A Reimann 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 approximating $\int_2^8 \frac{1}{tan(3^x)}dx$

So the interval is [2,8] , the subintervals have length $\frac{8-2}{2}=3$ , and since we are using the midpoints of each interval, the x-values are [3.5,6.5]

$\int_2^8 \frac{1}{tan(3^x)}dx \approx 3[\frac{1}{tan(3^{3.5})}+\frac{1}{tan(3^{6.5})}]=-19.5$

Report an Error

Example Question #96 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_3^9 \frac{3^{x}}{2^{x}}dx$ using three midpoints.

Possible Answers:

28.1

42.1

192.6

96.3

84.2

Correct answer:

84.2

Explanation:

A Reimann 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 approximating $\int_3^9 \frac{3^{x}}{2^{x}}dx$

So the interval is [3,9] , the subintervals have length $\frac{9-3}{3}=2$ , and since we are using the midpoints of each interval, the x-values are [4,6,8]

$\int_3^9 \frac{3^{x}}{2^{x}}dx \approx 2[\frac{3^{4}}{2^{4}}+\frac{3^{6}}{2^{6}}+\frac{3^{8}}{2^{8}}]=84.2$

Report an Error

Example Question #97 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_1^{2.5} tan(\frac{1}{x^{1.4}})$ using three midpoints.

Possible Answers:

0.348

0.861

0.696

1.722

0.132

Correct answer:

0.861

Explanation:

A Reimann 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 approximating $\int_1^{2.5} tan(\frac{1}{x^{1.4}})$

So the interval is [1,2.5] , the subintervals have length $\frac{2.5-1}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [1.25,1.75,2.25]

$\int_1^{2.5} tan(\frac{1}{x^{1.4}})\approx 0.5[tan(\frac{1}{1.25^{1.4}})+tan(\frac{1}{1.75^{1.4}})+tan(\frac{1}{2.25^{1.4}})]$

$\int_1^{2.5} tan(\frac{1}{x^{1.4}})\approx 0.861$

Report an Error

Example Question #98 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx$ using four midpoints.

Possible Answers:

1.904

1.656

0.952

3.312

3.808

Correct answer:

1.904

Explanation:

A Reimann 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 approximating $\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx$

So the interval is [2.5,4.5] , the subintervals have length $\frac{4.5-2.5}{4}=0.5$ , and since we are using the midpoints of each interval, the x-values are [2.75,3.25,3.75,4.25]

$\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx \approx 0.5[\frac{cos(2.75^3)}{tan(2.75^4)}+\frac{cos(3.25^3)}{tan(3.25^4)}+\frac{cos(3.75^3)}{tan(3.75^4)}+\frac{cos(4.25^3)}{tan(4.25^4)}]$

$\int_{2.5}^{4.5} \frac{cos(x^3)}{tan(x^4)}dx \approx 1.904$

Report an Error

Example Question #99 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx$ using three midpoints.

Possible Answers:

0.229

0.782

-0.014

1.114

-1.083

Correct answer:

0.782

Explanation:

A Reimann 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 approximating $\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx$

So the interval is [10.5,13.5] , the subintervals have length $\frac{13.5-10.5}{3}=1$ , and since we are using the midpoints of each interval, the x-values are [11,12,13]

$\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx \approx 1[tan(e^{11^{1.5}})+tan(e^{12^{1.5}})+tan(e^{13^{1.5}})]$

$\int_{10.5}^{13.5} tan(e^{x^{1.5}})dx \approx 0.782$

Report an Error

Example Question #101 : How To Find Midpoint Riemann Sums

Using the method of midpoint Reimann sums, approximate the integral $\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx$ using three midpoints.

Possible Answers:

1.94

1.35

2.03

0.97

2.69

Correct answer:

1.35

Explanation:

A Reimann 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 approximating $\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx$

So the interval is [0.25,1.75] , the subintervals have length $\frac{1.75-0.25}{3}=0.5$ , and since we are using the midpoints of each interval, the x-values are [0.5,1,1.5]

$\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx \approx 0.5[(3^{0.5^2}-2^{0.5^3})+(3^{1^2}-2^{1^3})+(3^{1.5^2}-2^{1.5^3})]$

$\int_{0.25}^{1.75} (3^{x^2}-2^{x^3})dx \approx 1.35$

Report an Error

View Calculus Tutors

Sayantan
Certified Tutor

Western Michigan University, Bachelor of Science, Applied Mathematics.

View Calculus Tutors

Metu
Certified Tutor

University of Central Arkansas, Bachelor of Science, Chemistry.

View Calculus Tutors

Blossom
Certified Tutor

University of Maryland-Baltimore County, Bachelor of Science, Computer Science.

All Calculus 1 Resources

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

Popular Subjects

English Tutors in Phoenix, GMAT Tutors in Miami, GMAT Tutors in Boston, Computer Science Tutors in San Diego, ISEE Tutors in Seattle, Physics Tutors in San Francisco-Bay Area, GRE Tutors in Miami, ISEE Tutors in San Francisco-Bay Area, Reading Tutors in San Diego, SAT Tutors in Los Angeles

Popular Courses & Classes

GMAT Courses & Classes in New York City, SAT Courses & Classes in Philadelphia, GRE Courses & Classes in Phoenix, GRE Courses & Classes in New York City, GMAT Courses & Classes in Philadelphia, ISEE Courses & Classes in Phoenix, GRE Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Los Angeles, SAT Courses & Classes in New York City, SSAT Courses & Classes in Seattle

Popular Test Prep

GMAT Test Prep in Philadelphia, GMAT Test Prep in Atlanta, MCAT Test Prep in Phoenix, LSAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Dallas Fort Worth, ISEE Test Prep in New York City, SSAT Test Prep in Atlanta, GRE Test Prep in Dallas Fort Worth, GRE Test Prep in Phoenix, SSAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #92 : How To Find Midpoint Riemann Sums

Example Question #93 : How To Find Midpoint Riemann Sums

Example Question #94 : Midpoint Riemann Sums

Example Question #94 : How To Find Midpoint Riemann Sums

Example Question #91 : How To Find Midpoint Riemann Sums

Example Question #96 : How To Find Midpoint Riemann Sums

Example Question #97 : How To Find Midpoint Riemann Sums

Example Question #98 : How To Find Midpoint Riemann Sums

Example Question #99 : How To Find Midpoint Riemann Sums

Example Question #101 : How To Find Midpoint Riemann Sums

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: