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Calculus 2 : Integral Applications

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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:

170m

172m

174m

171m

173m

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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Example Question #26 : Applications In Physics

In 1D electromagnetism, $V=\int_{a}^{b}E(r) dr$ , where is voltage drop, and is the electric field. and and are arbitrary bounds.

In a capacitor, E(r) is just a constant. Find the voltage drop for a constant electric field with strength E(r)=14 from r:(0,14)

Possible Answers:

420

210

Correct answer:

210

Explanation:

We can simply plug in the values into our first equation by:

$V=\int_{0}^{15}14dr=14*15= 210$

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Example Question #27 : Applications In Physics

In Electrical Engineering, the Laplace Transform is a heavily used integral transform. It is given by:

$L\left \{ f\right \}=\int_{0}^{\infty}e^{-st}*f(t) dt$ , where is a number.

Determine the Laplace Transform of $f(t)=e^{it}$

Assume i-s<0

Possible Answers:

$\frac{s+i}{s-i}$

$\frac{1}{s-i}$

$\frac{1}{s+i}$

$-\frac{1}{s+i}$

Correct answer:

$\frac{1}{s-i}$

Explanation:

$L\left \{ f\right \}=\int_{0}^{\infty}e^{-st}*e^{it} dt$

$\int e^{t(-s+i)} dt= \frac{1}{i-s}e^{t(i-s)} +C =F(t)$

By the fundamental theorem of calculus and since i-s<0 :

$\int_{0}^{\infty}e^{-st}*e^{it} dt = F(\infty)-F(0)= \frac{1}{i-s}(e^{-\infty}-e^0)$

$L\left \{ f\right \}=\int_{0}^{\infty}e^{-st}*e^{it} dt=\frac{1}{s-i}$

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Example Question #28 : Applications In Physics

In Electrical Engineering, the Laplace Transform is a heavily used integral transform. It is given by:

$L\left \{ f\right \}=\int_{0}^{\infty}e^{-st}*f(t) dt$ , where is a number.

Determine the Laplace transform of f(t)=e^t

Assume s-1>0

Possible Answers:

$\infty$

$\frac{1}{1-s}$

$\frac{1}{s-1}$

$\frac{1}{1-s}*e^{-s}$

Correct answer:

$\frac{1}{s-1}$

Explanation:

By the given formula:

$L\left \{ f\right \}=\int_{0}^{\infty}e^{-st}*e^t dt$

$\int e^{-t(s-1)} dt = \frac{1}{1-s}*e^{-t(s-1)}=F(t)$

By the fundamental theorem of calculus and because s-1>0 :

$\int_{0}^{\infty} e^{-t(s-1)} dt = F(\infty)-F(0)=\frac{1}{1-s}*(e^{-\infty}-e^0)$

$=\frac{1}{s-1}$

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Example Question #29 : Applications In Physics

A particle's position is given by the following equation:

s=t^3-6t^2+9t

Here, represents the particle's displacement from its starting point after seconds. What is the particle's acceleration at t=4 ?

Possible Answers:

Correct answer:

Explanation:

The velocity of an object is defined as the derivative of its position with respect to time; in turn, the acceleration of an object is defined as the derivative of its velocity with respect to time. Since we've been given the equation defining the particle's position after seconds, to determine its acceleration we must take the second derivative of this equation.

The first derivative with respect to time is

s'(t)=v(t)=3t^2-12t+9 .

The second derivative with respect to time is

s''(t)=v'(t)=a(t)=6t-12 .

Now we simply substitute t=4 for the new equation for the particle's acceleration to yield

a(t)=6(4)-12=12

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Example Question #30 : Applications In Physics

Water flows into a certain pool at a rate of 180+3t gal/min for an hour, where is measured in minutes. Find the amount of water that flows into the pool during the first minutes.

Possible Answers:

4000gal.

3800gal.

5200gal.

4200 gal.

2800gal.

Correct answer:

4200 gal.

Explanation:

We can model the flow of water into the pool by a function of time; let f(t) be the amount of water in the pool at time . Then f'(t)=180+3t represents the rate at which water flows into the pool at time . By the net change theorem, evaluating the definite integral of f'(t) over an interval [a,b] yields the total change, or amount of, that quantity over that interval. Hence, we can determine how much water flowed into the pool over the first minutes by evaluating the definite integral of f'(t) from to , as follows:

$\int_{0}^{20}f'(t)dt = \int_{0}^{20} (180+3t)dt$

$= \left.\180t+\frac{3}{2}t^2\right|_0^2^0$

=(3600+600)-(0+0)

=4200

Hence, the amount of water that flows into the pool during the first minutes at this rate is 4200gal. .

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Calculus 2 : Integral Applications

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #26 : Applications In Physics

Example Question #27 : Applications In Physics

Example Question #28 : Applications In Physics

Example Question #29 : Applications In Physics

Example Question #30 : Applications In Physics

All Calculus 2 Resources

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