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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 #3 : Applications In Physics

The velocity of a rocket, in meters per second, seconds after it was launched is modeled by

$v(t)=2\sqrt{t}$ .

What is the total distance travelled by the rocket during the first four seconds of its launch?

Possible Answers:

$32\ meters$

$\frac{16}{3}\ meters$

$16\ meters$

$\frac{32}{3}\ meters$

Correct answer:

$\frac{32}{3}\ meters$

Explanation:

The distance traveled by an object over an interval of time is the total area under the velocity curved during the interval. Therefore, we need to evaluate the definite integral

$\int_{0}^{4}2\sqrt{t}dt=\int_{0}^{4}2t^\frac{1}{2}dt$

Rewriting expressions written using radical notation using fractional exponents can make applying the power rule to find the antiderivative easier.

Evaluating this integral using the power rule,

$\int t^n=\frac{t^{n+1}}{n+1}+c$ , we find:

$\int_{0}^{4}2t^\frac{1}{2}dt=\frac{2}{3}\cdot 2(4^\frac{3}{2}) -\frac{2}{3}\cdot 2(0^\frac{3}{2})$

$=\frac{4}{3}\cdot (8)=\frac{32}{3}\ meters$

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Example Question #1 : Integral Applications

A force of

$F(x)=2x^2+3\ Newtons$

is applied to an object. How much work is done, in Joules, moving the object from x=1 to x=4 meters?

Possible Answers:

$\frac{5}{3} Joules$

$\frac{164}{3} Joules$

51Joules

$\frac{11}{3} Joules$

Correct answer:

51Joules

Explanation:

The work done moving an object is the intergral of force applied as a function of distance,

$W(x)=\int F(x)dx$ .

To find the work done moving the object from x=1 to x=4, evaluate the definite interval using x=1 and x=4 as the limits of integration. Recall that the antiderivative of a polynomial is found using the power rule,

$\int x^n=\frac{nx^{n+1}}{n+1}+c$ .

$\int_{1}^{4}(2x^2+3)dx=\frac{2(4^3)}{3}+3(4)-\frac{2(1^3)}{3}+3(1)$

$=\frac{164}{3}-\frac{11}{3}=\frac{153}{3}=51 joules$

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Example Question #1 : Applications In Physics

The velocity of an moviing object at time t=0 is v(0)=1 , and the position at time t=0 is s(0)=0 . The acceleration of the object is modeled by the function a(t)=4t for $t\ge0$ .

Possible Answers:

$s(t)={4t^2}+1$

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

s(t)=0

s(t)=4t+1

Correct answer:

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

Explanation:

The position function is the second antiderivative of acceleration. First, integrate the acceration function to find the velocity function. Since the acceleration function is a polynomial, the power rule is needed. Recall that the power rule is

$\int{x^ndx}=\frac{x^{n+1}}{n+1}+c$ .

$v(t)=\int{4t}dt=\frac{4t^2}{2}+c=2t^2+c$ .

Next, we can find the the value of the constant using the fact that v(0)=1 .

v(0)=1=2(0^2)+c = c , so c = 1 . The velocity function is therefore v(t)=2t^2+1 .

Now integrate the velocity function to find the position function.

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

Use the fact that s(0)=0 to find the constant .

$s(0)=0=\frac{2(0^3)}{3}+0+c=c,\rightarrow c=0$ .

The velocity function is

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

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Example Question #1 : Application Of Integrals

The temperature of an oven is increasing at a rate $r(t)=2e^{0.5t}$ degrees Fahrenheit per miniute for $0\le t\le10$ minutes. The initial temperature of the oven is T(0)=75 degrees Fahrenheit.

What is the temperture of the oven at t = 5 ? Round your answer to the nearest tenth.

Possible Answers:

323.9

212.2

119.7

239.4

Correct answer:

119.7

Explanation:

Integrating r(t) over an interval [a,b] will tell us the total accumulation, or change, in temperature over that interval. Therefore, we will need to evaluate the integral

$\int_{0}^{5}2e^{0.5t}dt$

to find the change in temperature that occurs during the first five minutes.

A substitution is useful in this case. Let $u=0.5t, du=0.5dt, \ and\ 2du=dt$ . We should also express the limits of integration in terms of . When t = 0, u=0 , and when t=5, u=2.5. Making these substitutions leads to the integral

$\int_{0}^{2.5}4e^udu$ .

To evaluate this, you must know the antiderivative of an exponential function.

In general,

$\int e^xdx=e^x+c$ .

Therefore,

$\int_{0}^{2.5}4e^udu=4e^u|_{0}^{2.5}=4e^{2.5}-4\approx 44.7$ .

This tells us that the temperature rose by approximately 44.7 degrees during the first five minutes. The last step is to add the initial temperature, which tells us that the temperature at t=5 minutes is

75+44.7=119.7 degrees.

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Example Question #1 : Integral Applications

Find the equation for the velocity of a particle if the acceleration of the particle is given by:

a(t)=16t^2+48

and the velocity at time t=0 of the particle is .

Possible Answers:

16t^2+48

$\frac{16}{3}t^3+96$

$\frac{16}{3}t^3+48t+30$

16t^2+78

Correct answer:

$\frac{16}{3}t^3+48t+30$

Explanation:

In order to find the velocity function, we must integrate the accleration function:

$\int (16t^2+48)dt=\frac{16}{3}t^3+48t+C$

We used the rule

$\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

to integrate.

Now, we use the initial condition for the velocity function to solve for C. We were told that

v(0)=30

so we plug in zero into the velocity function and solve for C:

$v(0)=30=\frac{16}{3}(0)^3+48(0)+C$

C is therefore 30.

Finally, we write out the velocity function, with the integer replacing C:

$v(t)=\frac{16}{3}t^3+48t+30$

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Example Question #1 : Initial Conditions

Find the work done by gravity exerting an acceleration of $-10\frac{m}{s^2}$ for a 10kg block down from its original position with no initial velocity.

Remember that

$F_{grav}=mass*acceleration$

$W=\int_{a}^{b}F(x)dx$ , where F(x) is a force measured in Newtons , is work measured in Joules , and and are initial and final positions respectively.

Possible Answers:

1000J

-500J

100J

250J

Correct answer:

250J

Explanation:

The force of gravity is proportional to the mass of the object and acceleration of the object.

$F=ma= (10*-10)\frac{kg*m}{s^2}= -100N$

Since the block fell down 5 meters, its final position is and initial position is .

$\\ W=\int_{0}^{-5}-100dx\\ \\=\frac{-100x}{2}|_{0}^{-5} \\ \\=-50x|_{0}^{-5}\\ \\=-50(-5)--50(0)\\ \\=250J$

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Example Question #1931 : Calculus Ii

The velocity of a car is defined by the equation $v(t)=4x^{2}-3x$ , where is the time in minutes. What distance (in meters) does the car travel between t=0 and t=4 ?

Possible Answers:

600m

60 m

$\frac{3}{184} m$

$\frac{184}{3} m$

$\frac{190}{3}m$

Correct answer:

$\frac{184}{3} m$

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since $v(t)=4x^{2}-3x$ , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{4}(4x^{2}-3x)dx$

$=\left[\frac{4}{3}x^{3}-\frac{3}{2}x^{2}\right]_{0}^{4}\textrm{}$

$=\left[\frac{4}{3}(4)^{3}-\frac{3}{2}(4)^{2}\right]-\left[\frac{4}{3}(0)^{3}-\frac{3}{2}(0)^{2}\right]$

$=\left[\frac{256}{3}-\frac{48}{2}\right]$

$=\frac{512}{6}-\frac{144}{6}$

$=\frac{368}{6}$

$=\frac{184}{3}$

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Example Question #11 : Integral Applications

The velocity of a train is defined by the equation $v(t)=x^{2}+6x-5$ , where is the time in seconds What distance (in meters) does the train travel between t=0 and t=3 ?

Possible Answers:

20m

21 m

10m

12m

25m

Correct answer:

21 m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since $v(t)=x^{2}+6x-5$ , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{3}(x^{2}+6x-5)dx$

$=\left[\frac{1}{3}x^{3}+3x^{2}-5x\right]_{0}^{3}\textrm{}$

$=\left[\frac{1}{3}(3)^{3}+3(3)^{2}-5(3)\right]-\left[\frac{1}{3}(0)^{3}+3(0)^{2}-5(0)\right]$

$=\frac{1}{3}(3)^{3}+3(3)^{2}-5(3)$

$=\frac{27}{3}+27-15$

=9+27-15

=36-15

=21

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Example Question #13 : Integral Applications

The velocity of a balloon is defined by the equation $v(t)=6x^{2}-4x+12$ , where is the time in minutes. What distance (in meters) does the balloon travel between t=0 and t=2 ?

Possible Answers:

34m

32m

30m

23m

33m

Correct answer:

32m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or $\int_{a}^{b}v(t)dt$ .

Since $v(t)=6x^{2}-4x+12$ , we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{2}(6x^{2}-4x+12)dx$

$=[2x^{3}-2x^{2}+12x]_{0}^{2}\textrm{}$

$=[2(2)^{3}-2(2)^{2}+12(2)]-[2(0)^{3}-2x=(0)^{2}+12(0)]$

=16-8+24

=8+24

=32

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Example Question #14 : Integral Applications

A frisbee has a velocity defined by v(t)=2t+13 , where we express in seconds. What distance does it travel between $0\leq t\leq 2$ in meters?

Possible Answers:

30m

32m

34m

28m

26m

Correct answer:

30m

Explanation:

We define velocity as the derivative of distance, or v(t)=d'(t) .

Thus, in order to find the distance traveled when given the velocity, we'll need to take the definite integral of the velocity function over a specified time period, or

$\int_{a}^{b}v(t)dt$ .

Since

v(t)=2t+13 ,

we can use the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{2}(2t+13)dt$

$=[t^{2}+13t]_{0}^{2}\textrm{}$

$=[(2)^{2}+13(2)]-[(0)^{2}+(0)t]$

=4+26

=30

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3 : Applications In Physics

Example Question #1 : Integral Applications

Example Question #1 : Applications In Physics

Example Question #1 : Application Of Integrals

Example Question #1 : Integral Applications

Example Question #1 : Initial Conditions

Example Question #1931 : Calculus Ii

Example Question #11 : Integral Applications

Example Question #13 : Integral Applications

Example Question #14 : Integral Applications

All Calculus 2 Resources

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