Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #121 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{5}^{16.6}(tan(12sin(6x)))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

124.00

4.04

78.79

12.92

Correct answer:

12.92

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{16.6}(tan(12sin(6x)))dx\\&\text{So the interval is }[5,16.6]\text{ the subintervals have length }\frac{16.6-(5)}{4}=\frac{29}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{129}{20},\frac{187}{20},\frac{49}{4},\frac{303}{20}]\\&\int_{5}^{16.6}(tan(12sin(6x)))dx=\frac{29}{10}[(tan(12sin(\frac{387}{10})))+(tan(12sin(\frac{561}{10})))+(tan(12sin(\frac{147}{2})))+(tan(12sin(\frac{909}{10})))]\\&\int_{5}^{16.6}(tan(12sin(6x)))dx=12.92\end{align*}$

Report an Error

Example Question #122 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

20.39

12.38

7.77

2.43

Correct answer:

2.43

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx\\&\text{So the interval is }[-4,-3.6]\text{ the subintervals have length }\frac{-3.6-(-4)}{4}=\frac{1}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{79}{20},-\frac{77}{20},-\frac{15}{4},-\frac{73}{20}]\\&\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx=\frac{1}{10}[(13sin(10tan(\frac{79}{10})))+(13sin(10tan(\frac{77}{10})))+(13sin(10tan(\frac{15}{2})))+(13sin(10tan(\frac{73}{10})))]\\&\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx=2.43\end{align*}$

Report an Error

Example Question #123 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{23.8}(3ln(-16tan(2x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

187.14

1048.00

21.03

729.86

Correct answer:

187.14

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{23.8}(3ln(-16tan(2x)))dx\\&\text{So the interval is }[5,23.8]\text{ the subintervals have length }\frac{23.8-(5)}{4}=\frac{47}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{147}{20},\frac{241}{20},\frac{67}{4},\frac{429}{20}]\\&\int_{5}^{23.8}(3ln(-16tan(2x)))dx=\frac{47}{10}[(3ln(-16tan(\frac{147}{10})))+(3ln(-16tan(\frac{241}{10})))+(3ln(-16tan(\frac{67}{2})))+(3ln(-16tan(\frac{429}{10})))]\\&\int_{5}^{23.8}(3ln(-16tan(2x)))dx=187.14\end{align*}$

Report an Error

Example Question #124 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

-18.43

-405.43

-6.65

-40.54

Correct answer:

-40.54

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx\\&\text{So the interval is }[4,15.2]\text{ the subintervals have length }\frac{15.2-(4)}{4}=\frac{14}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{27}{5},\frac{41}{5},11,\frac{69}{5}]\\&\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx=\frac{14}{5}[(-7sin(9e^{(\frac{54}{5})}))+(-7sin(9e^{(\frac{82}{5})}))+(-7sin(9e^{(22)}))+(-7sin(9e^{(\frac{138}{5})}))]\\&\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx=-40.54\end{align*}$

Report an Error

Example Question #125 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx\\&\text{Using midpoints over }3\text{ intervals.}\end{align*}$

Possible Answers:

-14.08

-325.29

-49.29

-5.24

Correct answer:

-49.29

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx\\&\text{So the interval is }[-2,10.6]\text{ the subintervals have length }\frac{10.6-(-2)}{3}=\frac{21}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{1}{10},\frac{43}{10},\frac{17}{2}]\\&\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx=\frac{21}{5}[(-13cos(15e^{(\frac{1}{10})}))+(-13cos(15e^{(\frac{43}{10})}))+(-13cos(15e^{(\frac{17}{2})}))]\\&\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx=-49.29\end{align*}$

Report an Error

Example Question #126 : Introduction To Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{-2}^{10}(-10sin(15sin(x)))dx\end{align*}$

Possible Answers:

-346.78

-68.00

-8.39

-18.89

Correct answer:

-68.00

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{10}(-10sin(15sin(x)))dx\\&\text{So the interval is }[-2,10]\text{ the subintervals have length }\frac{10-(-2)}{3}=4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[0,4,8]\\&\int_{-2}^{10}(-10sin(15sin(x)))dx=4[(0)+(-10sin(15sin(4)))+(-10sin(15sin(8)))]\\&\int_{-2}^{10}(-10sin(15sin(x)))dx=-68.00\end{align*}$

Report an Error

Example Question #127 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{1}^{11.5}(5cos(8cos(x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

1.21

10.02

63.10

31.05

Correct answer:

10.02

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{11.5}(5cos(8cos(x)))dx\\&\text{So the interval is }[1,11.5]\text{ the subintervals have length }\frac{11.5-(1)}{3}=\frac{7}{2}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{11}{4},\frac{25}{4},\frac{39}{4}]\\&\int_{1}^{11.5}(5cos(8cos(x)))dx=\frac{7}{2}[(5cos(8cos(\frac{11}{4})))+(5cos(8cos(\frac{25}{4})))+(5cos(8cos(\frac{39}{4})))]\\&\int_{1}^{11.5}(5cos(8cos(x)))dx=10.02\end{align*}$

Report an Error

Example Question #128 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{7.4}(8cos(6e^{(3x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-26.13

-2.21

-40.74

-4.43

Correct answer:

-4.43

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{7.4}(8cos(6e^{(3x)}))dx\\&\text{So the interval is }[5,7.4]\text{ the subintervals have length }\frac{7.4-(5)}{4}=\frac{3}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{53}{10},\frac{59}{10},\frac{13}{2},\frac{71}{10}]\\&\int_{5}^{7.4}(8cos(6e^{(3x)}))dx=\frac{3}{5}[(8cos(6e^{(\frac{159}{10})}))+(8cos(6e^{(\frac{177}{10})}))+(8cos(6e^{(\frac{39}{2})}))+(8cos(6e^{(\frac{213}{10})}))]\\&\int_{5}^{7.4}(8cos(6e^{(3x)}))dx=-4.43\end{align*}$

Report an Error

Example Question #129 : Introduction To Integrals

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-4}^{10}(-14ln(2x^{2}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-4794.40

-3683.50

-1812.52

-584.68

Correct answer:

-584.68

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{10}(-14ln(2x^{2}))dx\\&\text{So the interval is }[-4,10]\text{ the subintervals have length }\frac{10-(-4)}{4}=\frac{7}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{1}{2},3,\frac{13}{2},10]\\&\int_{-4}^{10}(-14ln(2x^{2}))dx=\frac{7}{2}[(14ln(2))+(-14ln(18))+(-14ln(\frac{169}{2}))+(-14ln(200))]\\&\int_{-4}^{10}(-14ln(2x^{2}))dx=-584.68\end{align*}$

Report an Error

Example Question #81 : Riemann Sums

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-2}^{1}(12cos(5e^{(6x)}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

27.05

246.20

64.93

3.92

Correct answer:

27.05

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{1}(12cos(5e^{(6x)}))dx\\&\text{So the interval is }[-2,1]\text{ the subintervals have length }\frac{1-(-2)}{3}=1\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-1,0,1]\\&\int_{-2}^{1}(12cos(5e^{(6x)}))dx=1[(12cos(5e^{(-6)}))+(12cos(5))+(12cos(5e^{(6)}))]\\&\int_{-2}^{1}(12cos(5e^{(6x)}))dx=27.05\end{align*}$

Report an Error

View Calculus 2 Tutors

VUSUMUZI
Certified Tutor

University at Buffalo, Master of Science, Medicinal and Pharmaceutical Chemistry. University at Buffalo, Master of Science, P...

View Calculus 2 Tutors

Sundeep
Certified Tutor

Indian Institute of Technology, Engineering in Computer Science, Electrical Engineering. New York University, Masters in Busi...

View Calculus 2 Tutors

Jerry
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Engineering Physics. University of Illinois at Chicago, Master of Sci...

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Boston, SSAT Tutors in Houston, LSAT Tutors in San Diego, SAT Tutors in Miami, MCAT Tutors in Philadelphia, Algebra Tutors in Seattle, LSAT Tutors in San Francisco-Bay Area, Computer Science Tutors in Miami, GRE Tutors in Miami, ACT Tutors in Chicago

Popular Courses & Classes

GMAT Courses & Classes in San Diego, GMAT Courses & Classes in Boston, Spanish Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Atlanta, MCAT Courses & Classes in Washington DC, SAT Courses & Classes in Atlanta, GMAT Courses & Classes in Denver, MCAT Courses & Classes in Chicago, SSAT Courses & Classes in Atlanta, SSAT Courses & Classes in San Diego

Popular Test Prep

GMAT Test Prep in Boston, GMAT Test Prep in San Diego, ISEE Test Prep in Denver, GMAT Test Prep in Seattle, LSAT Test Prep in New York City, LSAT Test Prep in Denver, MCAT Test Prep in Miami, ACT Test Prep in Houston, ISEE Test Prep in Philadelphia, MCAT Test Prep in New York City

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 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #121 : Introduction To Integrals

Example Question #122 : Introduction To Integrals

Example Question #123 : Introduction To Integrals

Example Question #124 : Introduction To Integrals

Example Question #125 : Introduction To Integrals

Example Question #126 : Introduction To Integrals

Example Question #127 : Introduction To Integrals

Example Question #128 : Introduction To Integrals

Example Question #129 : Introduction To Integrals

Example Question #81 : Riemann Sums

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: