Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Calculus II

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

Bertha
Certified Tutor

California Institute of Technology, Bachelor's (in progress), Chemical Engineering.

View Calculus 2 Tutors

George
Certified Tutor

Pensacola State College, Bachelor in Arts, Mathematics.

View Calculus 2 Tutors

Irina
Certified Tutor

Miami University-Oxford, Bachelor of Science, Botany. University of Pittsburgh-Pittsburgh Campus, Doctor of Philosophy, Mathe...

All Calculus 2 Resources

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

Popular Subjects

SSAT Tutors in Phoenix, Calculus Tutors in Houston, Biology Tutors in New York City, Math Tutors in Boston, MCAT Tutors in San Francisco-Bay Area, GRE Tutors in Chicago, ACT Tutors in Phoenix, Biology Tutors in Los Angeles, GRE Tutors in Miami, SAT Tutors in Seattle

Popular Courses & Classes

SSAT Courses & Classes in Boston, SAT Courses & Classes in Boston, LSAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Denver, GRE Courses & Classes in Denver, LSAT Courses & Classes in Seattle, SSAT Courses & Classes in San Diego, Spanish Courses & Classes in Phoenix, LSAT Courses & Classes in Dallas Fort Worth

Popular Test Prep

GRE Test Prep in Houston, ISEE Test Prep in Los Angeles, MCAT Test Prep in Phoenix, SAT Test Prep in Houston, MCAT Test Prep in Dallas Fort Worth, SSAT Test Prep in Seattle, GMAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Philadelphia, SSAT Test Prep in Boston, SSAT Test Prep in Los Angeles

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 : Calculus II

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: