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Calculus 2 : Integrals

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Example Questions

Example Question #15 : Integral Applications

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:

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

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

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

None of the above

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

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 #11 : Integral Applications

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:

128m $\displaystyle 128m$

130m $\displaystyle 130m$

122m $\displaystyle 122m$

126m $\displaystyle 126m$

124m $\displaystyle 124m$

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 #17 : Integral Applications

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:

56m $\displaystyle 56m$

50m $\displaystyle 50m$

45m $\displaystyle 45m$

46m $\displaystyle 46m$

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 #18 : Integral Applications

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:

55m $\displaystyle 55m$

59m $\displaystyle 59m$

58m $\displaystyle 58m$

56m $\displaystyle 56m$

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 $\displaystyle v(t)=14t-2$, 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 #11 : Integral Applications

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 $\displaystyle 61m$

57m $\displaystyle 57m$

58m $\displaystyle 58m$

59m $\displaystyle 59m$

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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Example Question #21 : Applications In Physics

The velocity of a ship is defined as $v(t)=\frac{1}{2}t+9$ $\displaystyle v(t)=\frac{1}{2}t+9$ (where time $\displaystyle t$ is measured in seconds). What distance (in meters) does the ship travel between $\displaystyle 0$ seconds and $\displaystyle 2$ seconds?

Possible Answers:

19m $\displaystyle 19m$

22m $\displaystyle 22m$

21m $\displaystyle 21m$

18m $\displaystyle 18m$

20m $\displaystyle 20m$

Correct answer:

19m $\displaystyle 19m$

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 $v(t)=\frac{1}{2}t+9$ $\displaystyle v(t)=\frac{1}{2}t+9$, we can use the Power Rule for Integrals

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

to find:

$\int_{0}^{2}(\frac{1}{2}t+9)dt$ $\displaystyle \int_{0}^{2}(\frac{1}{2}t+9)dt$

$=[\frac{1}{4}t^{2}+9t]_{0}^2{}\textrm{}$ $\displaystyle =[\frac{1}{4}t^{2}+9t]_{0}^2{}\textrm{}$

Since the definite integral at $\displaystyle 0$ is $\displaystyle 0$, we get:

$=\frac{1}{4}(2)^{2}+9(2)$ $\displaystyle =\frac{1}{4}(2)^{2}+9(2)$

=1+18 $\displaystyle =1+18$

=19 $\displaystyle =19$

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Example Question #22 : Applications In Physics

The velocity of a car is defined as $\displaystyle v(t)=5t+6$ (where time $\displaystyle t$ is measured in seconds). What distance (in meters) does the car travel between $\displaystyle 0$ seconds and $\displaystyle 3$ seconds?

Possible Answers:

$\frac{79}{2}m$ $\displaystyle \frac{79}{2}m$

$\frac{80}{2}m$ $\displaystyle \frac{80}{2}m$

$\frac{83}{2}m$ $\displaystyle \frac{83}{2}m$

$\frac{82}{2}m$ $\displaystyle \frac{82}{2}m$

$\frac{81}{2}m$ $\displaystyle \frac{81}{2}m$

Correct answer:

$\frac{81}{2}m$ $\displaystyle \frac{81}{2}m$

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 $\displaystyle v(t)=5t+6$, we can use the Power Rule for Integrals

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

to find:

$\int_{0}^{3}(5t+6)dt$ $\displaystyle \int_{0}^{3}(5t+6)dt$

$=[\frac{5}{2}t^{2}+6t]_{0}^{3}\textrm{}$ $\displaystyle =[\frac{5}{2}t^{2}+6t]_{0}^{3}\textrm{}$

Since the definite integral at $\displaystyle 0$ is $\displaystyle 0$, we get:

$=\frac{5}{2}(3)^{2}+6(3)$ $\displaystyle =\frac{5}{2}(3)^{2}+6(3)$

$=\frac{45}{2}+18$ $\displaystyle =\frac{45}{2}+18$

$=\frac{45}{2}+\frac{36}{2}$ $\displaystyle =\frac{45}{2}+\frac{36}{2}$

$=\frac{81}{2}$ $\displaystyle =\frac{81}{2}$

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Example Question #21 : Applications In Physics

The velocity of a rocket is defined as $\displaystyle v(t)=12t+7$ (where time $\displaystyle t$ is measured in seconds). What distance (in meters) does the rocket travel between $\displaystyle 1$ second and $\displaystyle 5$ seconds?

Possible Answers:

173m $\displaystyle 173m$

171m $\displaystyle 171m$

174m $\displaystyle 174m$

170m $\displaystyle 170m$

172m $\displaystyle 172m$

Correct answer:

172m $\displaystyle 172m$

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 $\displaystyle v(t)=12t+7$, we can use the Power Rule for Integrals

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

to find:

$\int_{1}^{5}(12t+7)dt$ $\displaystyle \int_{1}^{5}(12t+7)dt$

$=[6t^{2}+7t]_{1}^{5}\textrm{}$ $\displaystyle =[6t^{2}+7t]_{1}^{5}\textrm{}$

$=[6(5)^{2}+7(5)]-[6(1)^{2}+7(1)]$ $\displaystyle =[6(5)^{2}+7(5)]-[6(1)^{2}+7(1)]$

$\displaystyle =[185]-[13]$

=172 $\displaystyle =172$

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Example Question #24 : Applications In Physics

A dog travels a certain distance between $\displaystyle 0$ seconds and $\displaystyle 2$ seconds. If we define its velocity as $\displaystyle v(t)=6t-5$, what is that distance in meters?

Possible Answers:

$\displaystyle 5m$

$\displaystyle 3m$

$\displaystyle 2m$

$\displaystyle 1m$

$\displaystyle 4m$

Correct answer:

$\displaystyle 2m$

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 $\displaystyle v(t)=6t-5$, we can use the Power Rule for Integrals

$\int t^{n}dt=\frac{t^{n+1}}{n+1}$ $\displaystyle \int t^{n}dt=\frac{t^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{2}(6t-5)dt$ $\displaystyle \int_{0}^{2}(6t-5)dt$

$=[3t^{2}-5t]_{0}^{2}$ $\displaystyle =[3t^{2}-5t]_{0}^{2}$

$=[3(2)^{2}-5(2)]-[3(0)^{2}-5(0)]$ $\displaystyle =[3(2)^{2}-5(2)]-[3(0)^{2}-5(0)]$

$\displaystyle =[12-10]$

$\displaystyle =2$

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Example Question #25 : Applications In Physics

A skateboarder travels a certain distance between $\displaystyle 0$ seconds and $\displaystyle 4$ seconds. If we define her velocity as $\displaystyle v(t)=2t+1$, what is her distance in meters?

Possible Answers:

20m $\displaystyle 20m$

25m $\displaystyle 25m$

30m $\displaystyle 30m$

10m $\displaystyle 10m$

15m $\displaystyle 15m$

Correct answer:

20m $\displaystyle 20m$

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 $\displaystyle v(t)=2t+1$, we can use the Power Rule for Integrals

$\int t^{n}dt=\frac{t^{n+1}}{n+1}$ $\displaystyle \int t^{n}dt=\frac{t^{n+1}}{n+1}$ for all $n\neq0$ ,

to find:

$\int_{0}^{4}(2t+1)dt$ $\displaystyle \int_{0}^{4}(2t+1)dt$

$=[t^{2}+t]_{0}^{4}$ $\displaystyle =[t^{2}+t]_{0}^{4}$

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

=16+4 $\displaystyle =16+4$

=20 $\displaystyle =20$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #15 : Integral Applications

Example Question #11 : Integral Applications

Example Question #17 : Integral Applications

Example Question #18 : Integral Applications

Example Question #11 : Integral Applications

Example Question #21 : Applications In Physics

Example Question #22 : Applications In Physics

Example Question #21 : Applications In Physics

Example Question #24 : Applications In Physics

Example Question #25 : Applications In Physics

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