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Calculus 2 : Calculus II

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

Example Questions

Example Question #41 : Applications In Physics

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

Possible Answers:

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

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

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

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

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

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 #42 : Applications In Physics

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

Possible Answers:

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

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

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

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

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

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 #43 : Applications In Physics

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

Possible Answers:

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

x(t)=t^2+t

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

x(t)=t^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. 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

Example Question #51 : Integral Applications

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

Possible Answers:

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

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

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

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

x(t)=2t^2+t-17

Correct answer:

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

Explanation:

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

$\int 4t+1dt$

Now, integrate:
$\frac{4t^2}{2}+t=2t^2+t+C$

Now plug in your initial conditions:

2(2^2)+2+C=3; C=-7

Plug your C back into your position function:

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

Report an Error

Example Question #51 : Applications In Physics

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

Possible Answers:

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

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

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

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

x(t)=2t^2-t

Correct answer:

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

Explanation:

Recall that the integral of velocity is the position function:

$\int 4t-1dt$

Now, 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 to solve for C:

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

C=-3

Now plug your C back in to your position function:

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

Report an Error

Example Question #52 : Applications In Physics

If v(t)=6t+2 and x(1)=4 , what is the original position function?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$\int 6t+2dt$

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

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

Now, plug in your initial conditions:

3(1^2)+2(1)+C=4

C=-1

Now plug your C back into your position function:

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

Report an Error

Example Question #51 : Applications In Physics

In the vacuum of space, a particle is initially moving at a constant speed of only 6 meters/second away from you as you watch in your space capsule stationary relative to the particle. You track the object and then observe an acceleration that varies with time as,

$\small a(t)=2t^2$

After travelling for 20 minutes at this speed the particle is now $4\times 10^{9}$ km away from you. How far was it from you at the instant it began accelerating?

Possible Answers:

$x_o \approx 3.12\times 10^{9} m \: \: \:or\: \: \:3.12\times 10^{6} km$

$x_o \approx 2.12\times 10^{9} m \: \: \:or\: \: \:2.12\times 10^{6} km$

$x_o \approx 3.12\times 10^{12} m \: \: \:or\: \: \:3.12\times 10^{9} km$

$x_o \approx 3.65\times 10^{9} m \: \: \:or\: \: \:3.65\times 10^{6} km$

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Correct answer:

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Explanation:

$\small a(t)=2t^2$

To solve the problem we first need to find the position function by successively integrating. First find the velocity function by integrating the acceleration:

$\small v(t) = \int a(t)dt =\int 2t^2 dt =\frac{2}{3}t^3 +v_o$

$\small v(t) = \frac{2}{3}t^3 +v_o$

The constant of integration $\small v_o$ is the initial velocity, which we are given as 6 meters-per-second. For convenience, we will omit the units in the equation.

$\small v(t)=\frac{2}{3}t^3 +6$

Now we integrate the velocity function to find the position function:

$\small \small x(t)=\int v(t) dt = \int \left(\frac{2}{3}t^3 +6 \right )dt = \frac{2}{12}t^4+6t+x_o$

$\small \small \small x(t)= \frac{1}{6}t^4+6t+x_o$

The constant of integration x_o represents the initial position when t=0 . After $\small t= 20 minutes = 1200 seconds$ the particle is $4\times 10^{9}$ km, or $4\times 10^{12}$ meters, away. Using standard S.I. units (meters and seconds), solve for x_o at t=1200 seconds where $x(t=1200)=4\times 10^{12}$ .

$\small x (t=1200)\: \: =\: \: \frac{1}{6}(1200)^4+6(1200)+x_o\: \: \: =\: \: \: 4\times 10^{12}$

Now solve for x_o

$x_o = 4\times 10^{12} -\frac{1}{6}(1200)^4-6(1200)$

$x_o \approx 3.65\times 10^{12} m \: \: \:or\: \: \:3.65\times 10^{9} km$

Report an Error

Example Question #221 : Integrals

Give the area of the region bounded by the curves of the functions

f(x) =x^2-4x+7

and

g(x) = 2 x^2 -8x+10 .

Possible Answers:

$\frac{7}{6}$

$\frac{4}{3}$

$\frac{5}{6}$

$\frac{2}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

The two curves intersect when g(x) = f(x) .

2 x^2 -8x+10 = x^2-4x+7

2 x^2 - x^2 -8x+ 4x+10 - 7 = 0

x^2 -4x+ 3 = 0

(x-1)(x-3) = 0

x= 1,x=3

Therefore, the area between the curves is

$\left |\int_{1}^{3} \left [g(x) - f(x) \right ] \mathrm{d}x \right |$

$=\left |\int_{1}^{3} \left (x^2 -4x+ 3 \right) \mathrm{d}x \right |$

$=\left | \left.\begin{matrix} \frac{x^3}{3} -4\cdot \frac{x^{2}}{2}+ 3x \\ \\ \end{matrix}\right| \begin{matrix} 3\\ \\ 1 \end{matrix} \right |$

$=\left | \left.\begin{matrix} \frac{x^3}{3} -2 x^{2} + 3x \\ \\ \end{matrix}\right| \begin{matrix} 3\\ \\ 1 \end{matrix} \right |$

$=\left | \left (\frac{3^3}{3} -2 \cdot 3^{2} + 3 \cdot 3 \right ) - \left (\frac{1^3}{3} -2 \cdot 1^{2} + 3 \cdot 1 \right ) \right |$

$=\left | \left (9 -18 + 9 \right ) - \left (\frac{1}{3} -2 + 3 \right ) \right |$

$=\left | 0 - \frac{4}{3} \right | = \frac{4}{3}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Applications In Physics

Example Question #47 : Integral Applications

Example Question #42 : Applications In Physics

Example Question #43 : Applications In Physics

Example Question #50 : Integral Applications

Example Question #51 : Integral Applications

Example Question #51 : Applications In Physics

Example Question #52 : Applications In Physics

Example Question #51 : Applications In Physics

Example Question #221 : Integrals

All Calculus 2 Resources

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