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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.1.12b
Chapter 6, Problem 6.1.12b

Distance traveled and displacement Suppose an object moves along a line with velocity (in ft/s) v(t)=6−2t, for 0≤t≤6, where t is measured in seconds.


b. Find the displacement of the object on the interval 0≤t≤6.

Verified step by step guidance
1
Recall that displacement over a time interval is the net change in position, which can be found by integrating the velocity function over that interval.
Set up the definite integral for displacement using the given velocity function \(v(t) = 6 - 2t\) over the interval \(0 \leq t \leq 6\): \[\int_0^6 (6 - 2t) \, dt\]
Integrate the function \(6 - 2t\) with respect to \(t\). The integral of \(6\) is \$6t\(, and the integral of \)-2t\( is \)-t^2$.
Evaluate the antiderivative at the upper and lower limits of the interval, \(t=6\) and \(t=0\), and subtract to find the displacement: \[\left[6t - t^2\right]_0^6 = (6 \times 6 - 6^2) - (6 \times 0 - 0^2)\]
Simplify the expression to find the net displacement of the object over the time interval.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Velocity and Displacement

Velocity is the rate of change of position with respect to time and can be positive or negative, indicating direction. Displacement is the net change in position over a time interval and is found by integrating the velocity function over that interval.
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Using The Velocity Function

Definite Integral as Net Change

The definite integral of a velocity function from time a to b gives the net change in position, or displacement, during that time. It sums the signed areas under the velocity curve, accounting for direction.
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Definition of the Definite Integral

Evaluating Definite Integrals

To find displacement, compute the definite integral of v(t) from 0 to 6 by finding an antiderivative and applying the Fundamental Theorem of Calculus. This involves substituting the limits into the antiderivative and subtracting.
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Definition of the Definite Integral
Related Practice
Textbook Question

Determine whether the following statements are true and give an explanation or counterexample.


b. The area of the region between y=sin x and y=cos x on the interval [0,π/2] is ∫π/20(cosx−sinx)dx.

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Textbook Question

40–43. Population growth


Starting with an initial value of P(0)=55, the population of a prairie dog community grows at a rate of P′(t)=20−t/5 (prairie dogs/month), for 0≤t≤200, where t is measured in months.


b. Find the population P(t), for 0≤t≤200.

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Textbook Question

Two runners At noon (t=0), Alicia starts running along a long straight road at 4 mi/hr. Her velocity decreases according to the function v(t) = 4 / t + 1 for t≥0. At noon, Boris also starts running along the same road with a 2-mi head start on Alicia; his velocity is given by u(t) = 2 / t + 1, for t≥0. Assume t is measured in hours.


b. When, if ever, does Alicia overtake Boris?

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Textbook Question

Filling a tank A 2000-liter cistern is empty when water begins flowing into it (at t=0 at a rate (in L/min) given by Q′(t) = 3√t, where t is measured in minutes.


b. Find the function that gives the amount of water in the tank at any time t≥0.

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Textbook Question

Blood flow A typical human heart pumps 70 mL of blood (the stroke volume) with each beat. Assuming a heart rate of 60 beats/min (1 beat/s), a reasonable model for the outflow rate of the heart is V′(t)=70(1+sin 2πt), where V(t) is the amount of blood (in milliliters) pumped over the interval [0,t],V(0)=0 and t is measured in seconds.


b. Find the function that gives the total blood pumped between t=0 and a future time t>0.

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Textbook Question

55–58. Marginal cost Consider the following marginal cost functions.


b. Find the additional cost incurred in dollars when production is increased from 500 units to 550 units.


C′(x) = 300+10x−0.01x²

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