Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Integration

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : How To Find Position

Function v(t) gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models the particle's postion as a function of time.

Possible Answers:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}+\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}$

$\small \small h(t)=\frac{t^6}{5}+\frac{3t^5}{4}-\frac{7t^3}{9}+c$

$\small \small h(t)=\frac{t^6}{6}+\frac{3t^5}{7}-\frac{7t^3}{27}+c$

Correct answer:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Explanation:

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

We are starting with the following

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

We need to perform the following:

$\small \small \small \small\int v(t)dt=\int t^5+3t^4-\frac{7t^2}{3}dt$

Recall that to integrate, we add one to each exponent and divide by the that number, so we get the following. Don't forget your +c as well.

$\small \small \int t^5+3t^4-\frac{7t^2}{3}dt=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Which makes our position function, h(t), the following:

$\small h(t)=\frac{t^6}{6}+\frac{3t^5}{5}-\frac{7t^3}{9}+c$

Report an Error

Example Question #1 : How To Find Position

Consider the velocity function modeled in meters per second by v(t).

v(t)=4t^3-5t^2+6

Find the position of a particle whose velocity is modeled by v(t) after seconds.

Possible Answers:

$393.3\bar{3}m$

$8393.3\bar{3} m$

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

$93.3\bar{3}+c {m}$

$833.3\bar{3}+c m$

Correct answer:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

Explanation:

Recall that velocity is the first derivative of position, so to find the position function we need to integrate v(t) .

$V(t)=\int 4t^3-5t^2+6dt$

Becomes,

$V(t)=t^4-\frac{5t^3}{3}+6t+c$

Then, we need to find V(10)

$V(10)=10^4-\frac{5\cdot10^3}{3}+6\cdot10+c=8393.3\bar{3}+c$

So our final answer is:

$8393.3\bar{3}+c \text{ }\frac{m}{s}$

Report an Error

Example Question #1 : Integration

Find the position function given the velocity function:

v(t)=3t^5+4t

Possible Answers:

t^6+t^2

15t^4+4

3t^6+4t^2

$\frac{1}{2}t^6+\frac{1}{2}t^2$

$\frac{1}{2}t^6+2t^2$

Correct answer:

$\frac{1}{2}t^6+2t^2$

Explanation:

To find the position from the velocity function, integrate

v(t)=3t^5+4t by increasing the exponent of each t term and then dividing that term by the new exponent value.

$s(t)=\int v(t)dt= \frac{3}{6}t^6+2t^2=\frac{1}{2}t^6+2t^2$

Report an Error

Example Question #2 : Integration

Consider the velocity function given by v(t) :

v(t)=5t^4-343t^6+51t^2

Find the position of a particle after seconds if its velocity can be modeled by v(t) and the graph of its position function passes through the point (2,2) .

Possible Answers:

3828979

-3828.979

-38289.79

Correct answer:

Explanation:

Recall that velocity is the first derivative of position and acceleration is the second derivative of position. Therfore, we need to integrate v(t) to find p(t)

v(t)=5t^4-343t^6+51t^2

$p(t)=V(t)=\int 5t^4-343t^6+51t^2dt$

So we get:

p(t)=t^5-49t^7+17t^3+c

What we ultimately need is p(5), but first we need to find c: Use the point (2,2)

p(t)=t^5-49t^7+17t^3+c

$2=2^5-49\cdot 2^7+17\cdot 2^3+c$

c=6106

So our position function is:

p(t)=t^5-49t^7+17t^3+6106

$p(5)=5^5-49\cdot 5^7+17\cdot 5^3+6106=-3816769$

Report an Error

Example Question #3 : Integration

The velocity equation of an object is given by the equation v(t) = (4t^2)(3t) . What is the position of the object at time t = 4 if the initial position of the object is ?

Possible Answers:

768

192

Correct answer:

768

Explanation:

The position of the object can be found by integrating the velocity equation and solving for x(4) . To integrate the velocity equation we first rewrite the equation.

v(t) = (4t^2)(3t) = 12t^3

To integrate this equation we must use the power rule where,

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Applying this to the velocity equation gives us,

$x(t) = \int (12t^3)dt = \frac{12t^{3+1}}{3+1} = \frac{12t^4}{4} = 3t^4 +C$ .

We must solve for the value of by using the initial position of the object.

x(0) = 3(0)^4 + C =0 + C= 0

Therefore, C = 0 and x(t) = 3t^4 .

x(4) = 3(4)^4 = 3(256) = 768

Report an Error

Example Question #4 : Integration

The acceleration of an object is given by the equation a(t) = 4t+1 . What is the equation for the position of the object, if the object has an initial velocity of and an initial position of ?

Possible Answers:

x(t) = 2t^3 + t^2 + 10t

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 5t$

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

x(t) = 2t^3 - t^2 + 10t

Correct answer:

$x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Explanation:

To find the position of the object we must use the power rule to integrate the acceleration equation twice. The power rule is such that

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Therefore integrating the acceleration equation gives us

$v(t) = \int (4t+1)dt = \frac{4t^{1+1}}{1+1} +\frac{1t^{0+1}}{0+1} +C = 2t^2 + t +C$

We can solve for the value of by using the initial velocity of the object.

v(0) = 2(0)^2 +(0)+C = 10

Therefore C = 10 and v(t) = 2t^2 + t + 10

To find the position of the object we integrate the velocity equation.

$x(t) = \int (2t^2+t+10)dt = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} + \frac{10t^{0+1}}{0+1} +C = \frac{2t^3}{3}+\frac{t^2}{2}+10t +C$

We can solve for this new value of by using the object's initial position

$x(0) = \frac{2(0)^3}{3} + \frac{(0)^2}{2} + 10(0) +C = 0 + C = 0$

Therefore C = 0 and $x(t) = \frac{2t^3}{3} + \frac{t^2}{2} + 10t$

Report an Error

Example Question #5 : Integration

The velocity of an object is given by the equation v(t) = 6t -3 . What is the position of the object at time t= 4 if the object has a position of and time t = 1 ?

Possible Answers:

Correct answer:

Explanation:

To find the position of the object we must first find the position equation of the object. The position equation can be found by integrating the velocity equation. This can be done using the power rule where if

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$

Using this rule we find that

$x(t) = \int (6t-3)dt = \frac{6t^{1+1}}{1+1} - \frac{3t^{0+1}}{0+1} +C = 3t^2 - 3t +C$

Using the position of the object at time t = 1 we can solve for

x(1) = 3(1)^2 - 3(1) + C = 3 - 3 + C =0+C= 6

Therefore C = 6 and x(t) = 3t^2 - 3t +6

We can now find the position at time t = 4 .

x(4) = 3(4)^2 - 3(4) +6 = 48-12 +6 = 42

Report an Error

Example Question #31 : How To Find Position

The velocity of an object is v(t) = 3t^2 . What is the position of the object if its initial position is 1000 ?

Possible Answers:

x(t) = 6t +1000

x(t) = t^3 +1000

$x(t) = \frac{t^4}{4} +1000$

x(t) = t^3 -1000

Correct answer:

x(t) = t^3 +1000

Explanation:

The position is the integral of the velocity. By integrating with the power rule we can find the object's position.

The power rule is where

$f'(x) = x^n \rightarrow f(x) = \frac{x^{n+1}}{n+1}$ .

Therefore the position of the object is

$x(t) = \int (3t^2)dt = \frac{3t^{(2+1)}}{2+1} +C = t^3 +C$ .

We can solve for the constant using the object's initial position.

x(0) = (0)^3 + C = 0+C =1000

Therefore C= 1000 and x(t) = t^3 +1000 .

Report an Error

Example Question #37 : How To Find Position

If the velocity function of a car is v(t)= 3cos(t) , what is the position when t=1 ?

Possible Answers:

3sin(1)

$-\frac{9\pi}{2}$

$\frac{3\pi}{2}$

3cos(1)

Correct answer:

3sin(1)

Explanation:

To find the position from the velocity function, take the integral of the velocity function.

$\int v(t)\:dt=\int 3cos(t)\:dt$

s(t)=3sin(t)

Substitute t=1 .

s(t=1)=3sin(1)

Report an Error

Example Question #1 : Integration

If the velocity of a particle is $\small v(t)=t^3-3t^2$ , and its position at $\small \small t=2$ is $\small \small \small s(2)=4$ , what is its position at $\small \small t=4$ ?

Possible Answers:

$\small \small s(4)=4$

$\small s(4)=8$

$\small \small s(4)=0$

$\small \small s(4)=-4$

Correct answer:

$\small s(4)=8$

Explanation:

This problem can be done using the fundamental theorem of calculus. We hvae

$\small \small \int_2^4 v(t)dt=\int_2^4 s'(t)dt=s(4)-s(2)$

where $\small s(t)$ is the position at time $\small t$ . So we have

$\small \small \small s(4)=s(2)+\int_2^4 v(t)dt$

so we need to solve for $\small \small \small \small \int_2^4 v(t)dt$ :

$\small \small \small \small \small \small \int_2^4 v(t)dt=\int_2^4 (t^3-3t^2)dt=\frac{t^4}{4}-t^3|_2^4$

$\small =\left(\frac{4^4}{4}-4^3\right)-\left(\frac{2^4}{4}-2^3\right)=(4^3-4^3)-\left(\frac{16}{4}-8\right)=-(4-8)=4$

So then the position at $\small t=4$ is

$\small \small s(4)=s(2)+\int_2^4 v(t)dt=4+4=8$

Report an Error

View Calculus 3 Tutors

Yosef
Certified Tutor

Cooper Union for the Advancement of Science and Art, Bachelor of Engineering, Mechanical Engineering. Rensselaer Polytechnic ...

View Calculus 3 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Calculus 3 Tutors

Edem
Certified Tutor

University of Ghana, Bachelor, Mathematics. African Institute of Mathematical Science, Master's/Graduate, Mathematics.

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in Denver, Math Tutors in Denver, Chemistry Tutors in San Francisco-Bay Area, Computer Science Tutors in Dallas Fort Worth, English Tutors in Seattle, SAT Tutors in San Francisco-Bay Area, Physics Tutors in Phoenix, Biology Tutors in Los Angeles, Calculus Tutors in Phoenix, Algebra Tutors in Philadelphia

Popular Courses & Classes

ACT Courses & Classes in Philadelphia, Spanish Courses & Classes in Washington DC, MCAT Courses & Classes in Phoenix, MCAT Courses & Classes in New York City, Spanish Courses & Classes in Chicago, GRE Courses & Classes in Miami, GMAT Courses & Classes in Phoenix, GRE Courses & Classes in Boston, GRE Courses & Classes in Atlanta, SSAT Courses & Classes in Denver

Popular Test Prep

ISEE Test Prep in Denver, ACT Test Prep in Philadelphia, GRE Test Prep in Atlanta, LSAT Test Prep in Denver, GMAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Phoenix, ACT Test Prep in Denver, ISEE Test Prep in Atlanta, SSAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Atlanta

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 3 : Integration

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3 : How To Find Position

Example Question #1 : How To Find Position

Example Question #1 : Integration

Example Question #2 : Integration

Example Question #3 : Integration

Example Question #4 : Integration

Example Question #5 : Integration

Example Question #31 : How To Find Position

Example Question #37 : How To Find Position

Example Question #1 : Integration

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: