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The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval.

$ v(t) = t^2 - 2t - 3 $, $ 2 \le t \le 4 $

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a) $\frac{2}{3}$b) 4

06:15

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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The velocity function (in …

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The acceleration function …

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01:41

A particle moves with acce…

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given velocity function V f t which is equal to t squared minus two t minus three. We want to find the displacement and the distance traveled by the particle Over the interval 2- four. For the first part we have s this is equal to The integral from 2 to 4 of VFT which is just The integral from 2 to 4 of He squared -2 T -3 Bt. And integrating, we have 1 3rd TQ two times t squared over two minus three T. This evaluated from 2 to 4 and simplifying, we have 1 30 cube minus t squared minus three T. This evaluated from 2 to 4 and when these four we have 1/3 times four. Race to the third power minus four squared minus three times four. This minus 20 is too, we have one third times two to the third, power minus two squared minus three times two. This gives us the value of to over three and so the displacement of the particle Is to 3rd m for part B B house distance. This is equal to the Integral from 2 to 4 of the absolute value of VF t d T. And this is equal to the integral from 2 to 4 of the absolute value of t squared minus two t minus three Bt. Now the absolute value of t squared minus two t minus three. This is equal to the absolute value of t minus three Times T Plus one. This is equal to positive t minus three Time's D Plus one. If t is in The interval 3-4 And you have negative of T -3 time's D plus one yes T is in 2- three. And so we rewrite this into the Integral from 2 to 3 of the negative of t squared minus two t minus three D. T. Plus we have The integral from 3 to 4 of t squared minus two t minus three DT. Now integrating we have negative of one third t cube minus we have to t squared over two, that's D squared minus three T. This evaluated from 2 to 3 plus We have 1/3 t to the third power minus t squared minus three. T. This evaluated from 3 to 4. Now when these three we have negative of one third times three to the third power minus three squared minus three times three. This -20 is too you have one third times two to the third, power minus two Squared -3 times two. And then this plus we have when these four that's 1/3 Times 4 to the 3rd. Power minus four squared minus three times four. This minus when T is three we have 1/3 times 3 to the third power minus three squared minus. We have three times three and this gives us Negative of we have 9 -9 minus nine minus. We have 8/3 -4 -6 and then plus we have 64/3 -16 -12 minus. We have 9 -9 -9, and so this gives us the value of to over three, and so the Distance traveled by the particle is 2/3 m.

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