Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Definite 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 #221 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx\end{align*}$

Possible Answers:

-278.82

-14.57

-42.25

-392.88

Correct answer:

-42.25

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division. Although we have two functions,}\\&\text{since they're being added, we integrate separately and combine}\end{align*}$

$\begin{align*}\\&\text{the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx=-\frac{ (13cos(3x))}{6}-\frac{ (76cos(x + 1))}{3}|_{28}^{31}\\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx=-\frac{ (13cos(3(31)))}{6}-\frac{ (76cos((31) + 1))}{3}-(-\frac{ (13cos(3(28)))}{6}-\frac{ (76cos((28) + 1))}{3})\\&\int_{28}^{31}(\frac{(13sin(3x))}{2}+\frac{ (76sin(x + 1))}{3})dx=-42.25\end{align*}$

Report an Error

Example Question #211 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx\end{align*}$

Possible Answers:

-4.86

-0.77

-0.52

-16.05

Correct answer:

-4.86

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx=-\frac{ (47cos(x + 1))}{7}-\frac{ 5}{(34x)}|_{-23}^{-18}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx=-\frac{ (47cos((-18) + 1))}{7}-\frac{ 5}{(34(-18))}-(-\frac{ (47cos((-23) + 1))}{7}-\frac{ 5}{(34(-23))})\\&\int_{-23}^{-18}(\frac{(47sin(x + 1))}{7}+\frac{ 5}{(34x^{2})})dx=-4.86\end{align*}$

Report an Error

Example Question #212 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx\end{align*}$

Possible Answers:

102.41

44.12

2494.72

286.75

Correct answer:

286.75

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx=\frac{(31x^{2})}{12}-\frac{ (3e^{(-2x)})}{8}|_{17}^{20}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx=\frac{(31(20)^{2})}{12}-\frac{ (3e^{(-2(20))})}{8}-(\frac{(31(17)^{2})}{12}-\frac{ (3e^{(-2(17))})}{8})\\&\int_{17}^{20}(\frac{(31x)}{6}+\frac{ (3e^{(-2x)})}{4})dx=286.75\end{align*}$

Report an Error

Example Question #213 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx\end{align*}$

Possible Answers:

-0.91

-8.01

-4.55

-3.09

Correct answer:

-0.91

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx=\frac{sin(4x)}{47}+\frac{ (10sin(x + 2))}{3}|_{-12}^{-8}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx=\frac{sin(4(-8))}{47}+\frac{ (10sin((-8) + 2))}{3}-(\frac{sin(4(-12))}{47}+\frac{ (10sin((-12) + 2))}{3})\\&\int_{-12}^{-8}(\frac{(4cos(4x))}{47}+\frac{ (10cos(x + 2))}{3})dx=-0.91\end{align*}$

Report an Error

Example Question #214 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx\end{align*}$

Possible Answers:

3.95

38.35

0.58

1.24

Correct answer:

3.95

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx=\frac{(2sin(4x))}{47}+\frac{ (79ln(x))}{5}|_{9}^{11.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx=\frac{(2sin(4(11.5)))}{47}+\frac{ (79ln((11.5)))}{5}-(\frac{(2sin(4(9)))}{47}+\frac{ (79ln((9)))}{5})\\&\int_{9}^{11.5}(\frac{(8cos(4x))}{47}+\frac{ 79}{(5x)})dx=3.95\end{align*}$

Report an Error

Example Question #215 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-14}^{-10.5}(\frac{(19cos(x + 2))}{10}+\frac{ (8sin(x + 1))}{57})dx\end{align*}$

Possible Answers:

-0.23

-7.03

-2.27

-0.41

Correct answer:

-2.27

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-14}^{-10.5}(\frac{(19cos(x + 2))}{10}+\frac{ (8sin(x + 1))}{57})dx=\frac{(19sin(x + 2))}{10}-\frac{ (8cos(x + 1))}{57}|_{-14}^{-10.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\frac{(19sin((-10.5) + 2))}{10}-\frac{ (8cos((-10.5) + 1))}{57}-(\frac{(19sin((-14) + 2))}{10}-\frac{ (8cos((-14) + 1))}{57})\\&-2.27\end{align*}$

Report an Error

Example Question #216 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{8}^{9.5}(\frac{(31cos(4x))}{2}+\frac{ (4e^{(-2x)})}{27})dx\end{align*}$

Possible Answers:

-0.11

-2.67

-0.18

-0.99

Correct answer:

-0.99

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{8}^{9.5}(\frac{(31cos(4x))}{2}+\frac{ (4e^{(-2x)})}{27})dx=\frac{(31sin(4x))}{8}-\frac{ (2e^{(-2x)})}{27}|_{8}^{9.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&\frac{(31sin(4(9.5)))}{8}-\frac{ (2e^{(-2(9.5))})}{27}-(\frac{(31sin(4(8)))}{8}-\frac{ (2e^{(-2(8))})}{27})\\&-0.99\end{align*}$

Report an Error

Example Question #217 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{6}^{9.5}(\frac{8}{x}+\frac{ (7x^{2})}{4})dx\end{align*}$

Possible Answers:

53.97

377.81

41.98

134.93

Correct answer:

377.81

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{6}^{9.5}(\frac{8}{x}+\frac{ (7x^{2})}{4})dx=8ln(x) +\frac{ (7x^{3})}{12}|_{6}^{9.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&8ln((9.5)) +\frac{ (7(9.5)^{3})}{12}-(8ln((6)) +\frac{ (7(6)^{3})}{12})\\&377.81\end{align*}$

Report an Error

Example Question #218 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{6}^{9}(\frac{sin(x + 1)}{2}+\frac{ 1}{(8x^{2})})dx\end{align*}$

Possible Answers:

0.15

8.03

0.80

0.22

Correct answer:

0.80

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{6}^{9}(\frac{sin(x + 1)}{2}+\frac{ 1}{(8x^{2})})dx=-\frac{ cos(x + 1)}{2}-\frac{ 1}{(8x)}|_{6}^{9}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=-\frac{ cos((9) + 1)}{2}-\frac{ 1}{(8(9))}-(-\frac{ cos((6) + 1)}{2}-\frac{ 1}{(8(6))})\\&=0.80\end{align*}$

Report an Error

Example Question #219 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{20}^{22.5}(\frac{(29sin(3x))}{12}+\frac{ 1}{(6\cdot 2^{(\frac{x}{2})})})dx\end{align*}$

Possible Answers:

-0.73

-0.13

-6.87

-0.25

Correct answer:

-0.73

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{20}^{22.5}(\frac{(29sin(3x))}{12}+\frac{ 1}{(6\cdot 2^{(\frac{x}{2})})})dx=-\frac{ (29cos(3x))}{36}-\frac{ 1}{(3\cdot 2^{(\frac{x}{2})}ln(2))}|_{20}^{22.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=-\frac{ (29cos(3(22.5)))}{36}-\frac{ 1}{(3\cdot 2^{(\frac{(22.5)}{2})}ln(2))}-(-\frac{ (29cos(3(20)))}{36}-\frac{ 1}{(3\cdot 2^{(\frac{(20)}{2})}ln(2))})\\&=-0.73\end{align*}$

Report an Error

View Calculus 2 Tutors

Gary
Certified Tutor

SUNY at Albany, Bachelor of Science, Economics. Pace University-New York, Masters in Business Administration, Analysis and Fu...

View Calculus 2 Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

View Calculus 2 Tutors

Mohammed
Certified Tutor

University of Maryland-College Park, Bachelor of Science, Electrical Engineering. University of Maryland-College Park, Master...

All Calculus 2 Resources

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

Popular Subjects

SAT Tutors in Los Angeles, Reading Tutors in Houston, Spanish Tutors in Los Angeles, Chemistry Tutors in Miami, Biology Tutors in Los Angeles, Physics Tutors in Los Angeles, Biology Tutors in Miami, Statistics Tutors in Houston, ACT Tutors in Miami, ISEE Tutors in Philadelphia

Popular Courses & Classes

GRE Courses & Classes in Seattle, MCAT Courses & Classes in Washington DC, ISEE Courses & Classes in Philadelphia, ACT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Phoenix, Spanish Courses & Classes in Denver, MCAT Courses & Classes in Phoenix, LSAT Courses & Classes in Boston, MCAT Courses & Classes in Boston

Popular Test Prep

GMAT Test Prep in Boston, MCAT Test Prep in New York City, SAT Test Prep in Seattle, LSAT Test Prep in Atlanta, SSAT Test Prep in Denver, GMAT Test Prep in Chicago, MCAT Test Prep in Miami, LSAT Test Prep in Miami, ACT Test Prep in Los Angeles, MCAT Test Prep in Washington DC

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 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #221 : Finding Integrals

Example Question #211 : Definite Integrals

Example Question #212 : Definite Integrals

Example Question #213 : Definite Integrals

Example Question #214 : Definite Integrals

Example Question #215 : Definite Integrals

Example Question #216 : Definite Integrals

Example Question #217 : Definite Integrals

Example Question #218 : Definite Integrals

Example Question #219 : Definite Integrals

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: