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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 6.1.13b
Chapter 6, Problem 6.1.13b

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.


b. Find the displacement over the given interval. 


v(t) = 3t²−6t on [0, 3]

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1
Identify the given velocity function: \(v(t) = 3t^{2} - 6t\) and the time interval \([0, 3]\) seconds.
Recall that displacement over a time interval \([a, b]\) is found by integrating the velocity function over that interval: \(\text{Displacement} = \int_{a}^{b} v(t) \, dt\).
Set up the definite integral for displacement: \(\int_{0}^{3} (3t^{2} - 6t) \, dt\).
Integrate the function term-by-term: the integral of \$3t^{2}\( is \)t^{3}\(, and the integral of \)-6t\( is \)-3t^{2}\(, so the antiderivative is \)t^{3} - 3t^{2}$.
Evaluate the antiderivative at the bounds and subtract: calculate \(\left[ t^{3} - 3t^{2} \right]_{0}^{3} = (3^{3} - 3 \times 3^{2}) - (0 - 0)\) to find the displacement.

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Key Concepts

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

Velocity and Displacement Relationship

Velocity represents the rate of change of position with respect to time. Displacement over a time interval is the net change in position, which can be found by integrating the velocity function over that interval. This integral sums the instantaneous velocities to give the overall change in position.
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Definite Integral as Net Area

The definite integral of a velocity function over a time interval calculates the net area between the velocity curve and the time axis. Positive areas correspond to movement in one direction, while negative areas represent movement in the opposite direction. The sum of these areas gives the total displacement.
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Evaluating Definite Integrals

To find displacement, compute the definite integral of v(t) from the start to end time. This involves finding the antiderivative of v(t), then applying the Fundamental Theorem of Calculus by subtracting the antiderivative evaluated at the lower limit from that at the upper limit.
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Related Practice
Textbook Question

Equal integrals Without evaluating integrals, explain the following equalities. (Hint: Draw pictures.)


b. ∫²₀(25−(x²+1)²) dx = 2∫₁⁵ y√y−1 dy

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Textbook Question

Different axes of revolution Suppose R is the region bounded by y=f(x) and y=g(x) on the interval [a, b], where f(x)≥g(x).


b. How is this formula changed if x0>b?

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Textbook Question

Calculating work for different springs Calculate the work required to stretch the following springs 0.4 m from their equilibrium positions. Assume Hooke’s law is obeyed.

b. A spring that requires 2 J of work to be stretched 0.1 m from its equilibrium position

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Textbook Question

Deceleration A car slows down with an acceleration of a(t) = −15 ft/s². Assume v(0)=60 ft/s,s(0)=0, and t is measured in seconds.


b. How far does the car travel in the time it takes to come to rest?

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal.

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Textbook Question

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


b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

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