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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 v(t)=3t-4 , where we express in seconds. What distance does it travel between $0\leq t\leq 3$ in meters?

Possible Answers:

$\frac{5}{2}m$

$\frac{5}{3}m$

$\frac{2}{3}m$

None of the above

$\frac{3}{2}m$

Correct answer:

$\frac{3}{2}m$

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

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:

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

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

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

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

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

$=\frac{3}{2}$

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Example Question #11 : Applications In Physics

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

Possible Answers:

128m

126m

124m

130m

122m

Correct answer:

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

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:

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

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

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

=88+36

=124

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Example Question #17 : Integral Applications

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

Possible Answers:

56m

50m

45m

46m

54m

Correct answer:

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 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:

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

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

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

=[24+30]

=54

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Example Question #18 : Integral Applications

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

Possible Answers:

55m

59m

58m

56m

57m

Correct answer:

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 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:

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

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

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

=[63-6]

=57

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Example Question #11 : Integral Applications

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

Possible Answers:

60m

61m

57m

58m

59m

Correct answer:

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 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:

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

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

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

=75-10-12+4

=57

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Example Question #21 : Applications In Physics

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

Possible Answers:

19m

22m

21m

18m

20m

Correct answer:

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

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

to find:

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

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

Since the definite integral at is , we get:

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

=1+18

=19

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Example Question #22 : Applications In Physics

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

Possible Answers:

$\frac{79}{2}m$

$\frac{80}{2}m$

$\frac{83}{2}m$

$\frac{82}{2}m$

$\frac{81}{2}m$

Correct answer:

$\frac{81}{2}m$

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 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:

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

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

Since the definite integral at is , we get:

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

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

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

$=\frac{81}{2}$

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Example Question #21 : Applications In Physics

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

Possible Answers:

173m

171m

174m

170m

172m

Correct answer:

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 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:

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

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

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

=[185]-[13]

=172

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Example Question #24 : Applications In Physics

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

Possible Answers:

Correct answer:

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)=6t-5 , we can use the Power Rule for Integrals

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

to find:

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

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

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

=[12-10]

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Example Question #25 : Applications In Physics

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

Possible Answers:

20m

25m

30m

10m

15m

Correct answer:

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

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

to find:

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

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

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

=16+4

=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 : Applications In Physics

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

All Calculus 2 Resources

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