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Calculus 2 : Integrals

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Example Questions

Example Question #36 : Applications In Physics

If v(t)=4t-1 and x(2)=2 , what is the original position function?

Possible Answers:

x(t)=-2t^2-t-4

x(t)=2t^2-2t-4

x(t)=2t^2-t-4

x(t)=2t^2-t+4

x(t)=2t^2-t

Correct answer:

x(t)=2t^2-t-4

Explanation:

First, integrate the velocity function to get the indefinite position function:

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

Now, plug in your initial conditions to solve for C:

2(2^2)-2+C=2; C=-4

Now, plug back into the position function:

x(t)=2t^2-t-4 .

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Example Question #37 : Applications In Physics

If v(t)=2t^2-1 and x(1)=2 , what is the original position function?

Possible Answers:

$x(t)=\frac{2}{3}t^2-t+\frac{7}{3}$

$x(t)=-\frac{2}{3}t^3-t+\frac{7}{3}$

$x(t)=\frac{2}{3}t^3-t+\frac{1}{3}$

$x(t)=\frac{2}{3}t^3-t+\frac{7}{3}$

$x(t)=\frac{2}{3}t^3-t+\frac{4}{3}$

Correct answer:

$x(t)=\frac{2}{3}t^3-t+\frac{7}{3}$

Explanation:

First, write the integral expression for this problem:

$\int 2t^2-1dt$

Now, integrate. Remember to add one to the exponent and then put that result on the denominator:

$2(\frac{t^3}{3})-t+C$

Now, plug in your initial conditions to find C:

$\frac{2(1)}{3}-1+C=2; C=\frac{7}{3}$

Now, plug your C into the position function:

$x(t)=\frac{2}{3}t^3-t+\frac{7}{3}$ .

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Example Question #38 : Applications In Physics

If v(t)=2t-1 and x(3)=4 , what is the original position function?

Possible Answers:

x(t)=t^2-t-2

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

x(t)=t^2-t

x(t)=2t^2-t-2

x(t)=t^2-4t-2

Correct answer:

x(t)=t^2-t-2

Explanation:

First, set up the integral expression:

$\int 2t-1dt$

Now, integrate. Remember to raise the exponent by 1 and put that result on the denominator. Also remember to add a C because it is an indefinite integral:

$\frac{2t^2}{2}-t+C=t^2-t+C$

Plug in your initial conditions to solve for C:

(3^2)-3+C=4; C=-2

Plug back in for C and get your position function:

x(t)=t^2-t-2 .

Report an Error

Example Question #39 : Applications In Physics

If v(t)=4t+2 and x(2)=1 , what is the original position function?

Possible Answers:

x(t)=2t^2+2t-32

x(t)=2t^2+2t-1

x(t)=t^2+2t-11

x(t)=2t^2+2t-11

x(t)=2t^2-2t-11

Correct answer:

x(t)=2t^2+2t-11

Explanation:

Remember that the integral of the velocity function is the position function. First, write the integral expression:

$\int 4t+2dt$

Integrate. Remember to raise the exponent by 1 and then put that result on the denominator:

$\frac{4t^2}{2}+2t=2t^2+2t+C$

Now, plug in your initial conditions to solve for C:

2(2^2)+2(2)+C=1; C=-11

Plug back in to get your position function:

x(t)=2t^2+2t-11

Report an Error

Example Question #40 : Applications In Physics

If v(t)=2t-1 and x(1)=2 , what is the original position function?

Possible Answers:

x(t)=t^2-t-2

x(t)=4t^2-t+2

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

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

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

Correct answer:

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

Explanation:

Recall that the integral of the velocity function is the position function. Therefore, set up the integral:
$\int 2t-1dt$

Now, integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

$\frac{2t^2}{2}-t=t^2-t+C$

Now, plug in your initial conditions to find C:
1^2-1+C=2; C=2

Plug back in to get your position function:

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

Report an Error

Example Question #211 : Integrals

If v(t)=6t-1 and x(1)=2 , what is the original position function?

Possible Answers:

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

x(t)=3t^2-t

x(t)=3t^2-t+4

x(t)=3t^2-t-6

x(t)=3t^2-t-8

Correct answer:

x(t)=3t^2-t

Explanation:

Recall that integrating the velocity function will yield you the position function:

$\int 6t-1dt$

Now, integrate:

$\frac{6t^2}{2}-t=3t^2-t+C$

Now, to find your C, plug in your initial conditions:

3(1^2)-1+C=2; C=0

Since, C is 0, your position function is:

x(t)=3t^2-t .

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Example Question #42 : Integral Applications

If v(t)=4t+5 and x(0)=3 , what is the original position function?

Possible Answers:

x(t)=t^2+5t+3

x(t)=2t^2+7t+3

x(t)=2t^2+5t-3

x(t)=2t^2+5t+3

x(t)=2t^2-5t+3

Correct answer:

x(t)=2t^2+5t+3

Explanation:

Recall that integrating the velocity function will yield the position function:

$\int 4t+5dt$

Now, integrate.

$\frac{4t^2}{2}+5t=2t^2+5t+C$

To find your C, plug in your initial conditions:

2(0^2)+0+C=3; C=3

Plug your C back in to get your position function:

x(t)=2t^2+5t+3

Report an Error

Example Question #43 : Integral Applications

If v(t)=4x^2-3x and x(1)=2 , what is the original position function?

Possible Answers:

$x(t)=\frac{4}{3}x^3-\frac{3}{2}x^2-\frac{13}{6}$

$x(t)=\frac{4}{3}x^3-\frac{3}{2}x^2+\frac{13}{6}$

$x(t)=\frac{4}{3}x^3-\frac{5}{2}x^2+\frac{13}{6}$

$x(t)=-\frac{4}{3}x^3-\frac{3}{2}x^2+\frac{13}{6}$

$x(t)=\frac{4}{3}x^2-\frac{3}{2}x+\frac{13}{6}$

Correct answer:

$x(t)=\frac{4}{3}x^3-\frac{3}{2}x^2+\frac{13}{6}$

Explanation:

First, set up the integral expression:

$\int 4x^2-3xdx$

Next, integrate:

$\frac{4x^3}{3}-\frac{3x^2}{2}+C$

Now, plug in your initial conditions:

$\frac{4}{3}(1^3)-\frac{3}{2}(1^2)+C=2; C=\frac{13}{6}$

Now, plug your C back into the position function:

$x(t)=\frac{4}{3}x^3-\frac{3}{2}x^2+\frac{13}{6}$

Report an Error

Example Question #44 : Integral Applications

If v(t)=10t-2 and x(1)=2 , what is the original position function?

Possible Answers:

x(t)=5t^2-2t-5

x(t)=5t^2-2t-1

x(t)=5t^2-2t+1

x(t)=3t^2-2t-1

x(t)=5t^2+2t-1

Correct answer:

x(t)=5t^2-2t-1

Explanation:

Recall that the integral of the velocity function is the position function:

$\int 10t-2dt$

Now, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\frac{10t^2}{2}-2t=5t^2-2t+C$

Now, plug in your initial conditions:

5(1^2)-2(1)+C=2; C=-1

Now, plug in your C to your position function:

x(t)=5t^2-2t-1

Report an Error

Example Question #45 : Integral Applications

If v(t)=2t-1 and x(1)=2 , what is the original position function?

Possible Answers:

x(t)=t^2-4t+2

x(t)=t^2-t-2

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

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

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

Correct answer:

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

Explanation:

Recall that the integral of the velocity function is the position function. Write the integral expression:

$\int 2t-1dt$

Now, integrate:

$\frac{2t^2}{2}-t=t^2-t+C$

Plug in your initial conditions to get your C:
1^2-1+C=2; C=2

Plug your C back into the position function:

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

Report an Error

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #36 : Applications In Physics

Example Question #37 : Applications In Physics

Example Question #38 : Applications In Physics

Example Question #39 : Applications In Physics

Example Question #40 : Applications In Physics

Example Question #211 : Integrals

Example Question #42 : Integral Applications

Example Question #43 : Integral Applications

Example Question #44 : Integral Applications

Example Question #45 : Integral Applications

All Calculus 2 Resources

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