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Calculus 2 : Integral Applications

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Example Questions

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Example Question #41 : Integral Applications

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

Possible Answers:

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

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

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

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

x(t)=3t^2-t

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 .

Report an Error

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

Example Question #46 : Integral Applications

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

Possible Answers:

x(t)=t^2+2t-23

x(t)=2t^2-2t-23

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

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

x(t)=2t^2+2t-23

Correct answer:

x(t)=2t^2+2t-23

Explanation:

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

$\int 4t+2dt$

Now integrate:

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

Plug in your initial conditions to find C:
2(3^2)+2(3)+C=1; C=-23

Plug your C into the position function:

x(t)=2t^2+2t-23

Report an Error

Example Question #47 : Integral Applications

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

Possible Answers:

x(t)=2t^2-t

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

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

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

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

Correct answer:

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

Explanation:

Recall that the integral of velocity is position. Therefore, write the integral expression:

$\int 4t-1dt$

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

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

Now, plug in your initial conditions:

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

Plug your C back in to your position function:

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

Report an Error

Example Question #48 : Integral Applications

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

Possible Answers:

x(t)=3t^2+4t-11

x(t)=t^2+4t-11

x(t)=t^2-4t-11

x(t)=t^2+4t+11

x(t)=t^2+4t-1

Correct answer:

x(t)=t^2+4t-11

Explanation:

Recall that the integral of velocity is position. Therefore, write your integral expression:

$\int 2t+4dt$

Now, integrate:

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

Plug in your initial conditions to find C:
2^2+4(2)+C=1; C=-11

Plug your C back into your position function:

x(t)=t^2+4t-11

Report an Error

Example Question #49 : Integral Applications

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

Possible Answers:

x(t)=t^2+t

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

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

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

x(t)=t^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, write your integral expression:

$\int 2t+1dt$

Integrate:

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

Plug in your initial conditions to find C:

1^2+1+C=4; C=2

Plug C back in to your position function:

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

Report an Error

Example Question #50 : Integral Applications

If v(t)=6t-1 and x(1)=3 , what is the original position function?

Possible Answers:

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

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

x(t)=3t^2-6t+1

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

x(t)=3t^2-4t+1

Correct answer:

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

Explanation:

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

$\int 6t-1dt$

Now, integrate:

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

Plug in your initial conditions to get your C:
3(1^2)-1+C=3; C=1

Plug your C back into the position function:

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

Report an Error

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Integral Applications

Example Question #42 : Integral Applications

Example Question #43 : Integral Applications

Example Question #44 : Integral Applications

Example Question #45 : Integral Applications

Example Question #46 : Integral Applications

Example Question #47 : Integral Applications

Example Question #48 : Integral Applications

Example Question #49 : Integral Applications

Example Question #50 : Integral Applications

All Calculus 2 Resources

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