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Calculus 2 : Integral Applications

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Example Questions

Example Question #3 : Initial Conditions

Find the work done by gravity exerting an acceleration of $-10\frac{m}{s^2}$ $\displaystyle -10\frac{m}{s^2}$ for a 10kg $\displaystyle 10kg$ block down $\displaystyle 5m$ from its original position with no initial velocity.

Remember that

$F_{grav}=mass*acceleration$ $\displaystyle F_{grav}=mass*acceleration$

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

Possible Answers:

250J $\displaystyle 250J$

-500J $\displaystyle -500J$

1000J $\displaystyle 1000J$

100J $\displaystyle 100J$

Correct answer:

250J $\displaystyle 250J$

Explanation:

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

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

Since the block fell down 5 meters, its final position is $\displaystyle -5$ and initial position is $\displaystyle 0$.

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

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Example Question #11 : Applications In Physics

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

Possible Answers:

600m $\displaystyle 600m$

$\frac{3}{184} m$ $\displaystyle \frac{3}{184} m$

$\frac{184}{3} m$ $\displaystyle \frac{184}{3} m$

$\frac{190}{3}m$ $\displaystyle \frac{190}{3}m$

60 m $\displaystyle 60 m$

Correct answer:

$\frac{184}{3} m$ $\displaystyle \frac{184}{3} m$

Explanation:

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

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

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

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

to find:

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

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

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

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

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

$=\frac{368}{6}$ $\displaystyle =\frac{368}{6}$

$=\frac{184}{3}$ $\displaystyle =\frac{184}{3}$

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Example Question #12 : Applications In Physics

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

Possible Answers:

12m $\displaystyle 12m$

25m $\displaystyle 25m$

21 m $\displaystyle 21 m$

10m $\displaystyle 10m$

20m $\displaystyle 20m$

Correct answer:

21 m $\displaystyle 21 m$

Explanation:

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

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

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

to find:

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

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

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

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

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

$\displaystyle =9+27-15$

=36-15 $\displaystyle =36-15$

=21 $\displaystyle =21$

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Example Question #13 : Applications In Physics

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

Possible Answers:

32m $\displaystyle 32m$

30m $\displaystyle 30m$

34m $\displaystyle 34m$

23m $\displaystyle 23m$

33m $\displaystyle 33m$

Correct answer:

32m $\displaystyle 32m$

Explanation:

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

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

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

to find:

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

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

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

$\displaystyle =16-8+24$

=8+24 $\displaystyle =8+24$

=32 $\displaystyle =32$

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Example Question #14 : Applications In Physics

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

Possible Answers:

26m $\displaystyle 26m$

30m $\displaystyle 30m$

32m $\displaystyle 32m$

28m $\displaystyle 28m$

34m $\displaystyle 34m$

Correct answer:

30m $\displaystyle 30m$

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

$\displaystyle v(t)=2t+13$,

we can use the Power Rule for Integrals

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

to find:

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

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

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

=4+26 $\displaystyle =4+26$

=30 $\displaystyle =30$

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Example Question #191 : Integrals

A ball has a velocity defined by $\displaystyle v(t)=3t-4$, where we express $\displaystyle t$ in seconds. What distance does it travel between $0\leq t\leq 3$ $\displaystyle 0\leq t\leq 3$ in meters?

Possible Answers:

$\frac{5}{2}m$ $\displaystyle \frac{5}{2}m$

$\frac{2}{3}m$ $\displaystyle \frac{2}{3}m$

$\frac{3}{2}m$ $\displaystyle \frac{3}{2}m$

None of the above

$\frac{5}{3}m$ $\displaystyle \frac{5}{3}m$

Correct answer:

$\frac{3}{2}m$ $\displaystyle \frac{3}{2}m$

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

$\displaystyle v(t)=3t-4$, we can use the Power Rule for Integrals

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

to find:

$\int_{0}^{3}(3t-4)dt$ $\displaystyle \int_{0}^{3}(3t-4)dt$

$=[\frac{3}{2}t^{2}-4t]_{0}^{3}$ $\displaystyle =[\frac{3}{2}t^{2}-4t]_{0}^{3}$

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

$=\frac{27}{2}-12$ $\displaystyle =\frac{27}{2}-12$

$=\frac{27}{2}-\frac{24}{2}$ $\displaystyle =\frac{27}{2}-\frac{24}{2}$

$=\frac{3}{2}$ $\displaystyle =\frac{3}{2}$

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Example Question #192 : Integrals

A subway has a velocity defined by $\displaystyle v(t)=11t+9$, where we express $\displaystyle t$ in seconds. What distance does it travel between $0\leq t\leq 4$ $\displaystyle 0\leq t\leq 4$ in meters?

Possible Answers:

126m $\displaystyle 126m$

128m $\displaystyle 128m$

122m $\displaystyle 122m$

124m $\displaystyle 124m$

130m $\displaystyle 130m$

Correct answer:

124m $\displaystyle 124m$

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

$\displaystyle v(t)=11t+9$,

we can use the Power Rule for Integrals

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

to find:

$\int_{0}^{4}(11t+9)dt$ $\displaystyle \int_{0}^{4}(11t+9)dt$

$=[\frac{11}{2}t^{2}+9t]_{0}^{4}\textrm{}$ $\displaystyle =[\frac{11}{2}t^{2}+9t]_{0}^{4}\textrm{}$

$=\frac{176}{2}+36$ $\displaystyle =\frac{176}{2}+36$

=88+36 $\displaystyle =88+36$

=124 $\displaystyle =124$

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Example Question #193 : Integrals

A train goes a certain distance between $0\leq t \leq 2$ $\displaystyle 0\leq t \leq 2$ (where $\displaystyle t$ is time in seconds). If we know that the train's velocity is defined as $\displaystyle v(t)=12t+15$, what is the distance it travelled (in meters)?

Possible Answers:

45m $\displaystyle 45m$

56m $\displaystyle 56m$

46m $\displaystyle 46m$

50m $\displaystyle 50m$

54m $\displaystyle 54m$

Correct answer:

54m $\displaystyle 54m$

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 $\displaystyle v(t)=12t+15$, we can use the Power Rule for Integrals

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

to find:

$\int_{0}^{2}(12t+15)dt$ $\displaystyle \int_{0}^{2}(12t+15)dt$

$=[6t^{2}+15t]_{0}^{2}\textrm{}$ $\displaystyle =[6t^{2}+15t]_{0}^{2}\textrm{}$

$=[6(2)^{2}+15(2)]-[6(0)^{2}+15(0)]$ $\displaystyle =[6(2)^{2}+15(2)]-[6(0)^{2}+15(0)]$

$\displaystyle =[24+30]$

=54 $\displaystyle =54$

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Example Question #194 : Integrals

A car goes a certain distance between $0\leq t \leq 3$ $\displaystyle 0\leq t \leq 3$ (where $\displaystyle t$ is time in seconds). If we know that the car's velocity is defined as $\displaystyle v(t)=14t-2$, what is the distance it travelled (in meters)?

Possible Answers:

59m $\displaystyle 59m$

57m $\displaystyle 57m$

55m $\displaystyle 55m$

56m $\displaystyle 56m$

58m $\displaystyle 58m$

Correct answer:

57m $\displaystyle 57m$

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 $\displaystyle v(t)=14t-2$, we can use the Power Rule for Integrals

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

to find:

$\int_{0}^{3}(14t-2)dt$ $\displaystyle \int_{0}^{3}(14t-2)dt$

$=[7t^{2}-2t]_{0}^{3}\textrm{}$ $\displaystyle =[7t^{2}-2t]_{0}^{3}\textrm{}$

$=[7(3)^{2}-2(3)]-[7(0)^{2}-2(0)]$ $\displaystyle =[7(3)^{2}-2(3)]-[7(0)^{2}-2(0)]$

=[63-6] $\displaystyle =[63-6]$

=57 $\displaystyle =57$

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Example Question #195 : Integrals

A plane goes a certain distance between $2\leq t \leq 5$ $\displaystyle 2\leq t \leq 5$ (where $\displaystyle t$ is time in seconds). If we know that the plane's velocity is defined as $\displaystyle v(t)=6t-2$, what is the distance it travelled (in meters)?

Possible Answers:

60m $\displaystyle 60m$

61m $\displaystyle 61m$

59m $\displaystyle 59m$

58m $\displaystyle 58m$

57m $\displaystyle 57m$

Correct answer:

57m $\displaystyle 57m$

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 $\displaystyle v(t)=6t-2$, we can use the Power Rule for Integrals

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

to find:

$\int_{2}^{5}(6t-2)dt$ $\displaystyle \int_{2}^{5}(6t-2)dt$

$=[3t^{2}-2t]_{2}^{5}\textrm{}$ $\displaystyle =[3t^{2}-2t]_{2}^{5}\textrm{}$

$=[3(5)^{2}-2(5)]-[3(2)^{2}-2(2)]$ $\displaystyle =[3(5)^{2}-2(5)]-[3(2)^{2}-2(2)]$

$\displaystyle =75-10-12+4$

=57 $\displaystyle =57$

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3 : Initial Conditions

Example Question #11 : Applications In Physics

Example Question #12 : Applications In Physics

Example Question #13 : Applications In Physics

Example Question #14 : Applications In Physics

Example Question #191 : Integrals

Example Question #192 : Integrals

Example Question #193 : Integrals

Example Question #194 : Integrals

Example Question #195 : Integrals

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