# The velocity function is v(t)= -(t^2)+6t-8for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-2,5].

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Luca B. | Certified Educator

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You need to remember that you may find the displacement evaluating the definite integral of velocity function, such that:

`d(t) = int_(-2)^5 (-(t^2) + 6t - 8) dt`

You need to use the property of linearity of integral such that:

`d(t) = int_(-2)^5 (-(t^2)) dt + int_(-2)^5 6t dt - int_(-2)^5 8dt`

You need to use the fundamental theorem of calculus, such that:

`d(t) = -t^3/3|_(-2)^5 + 6t^2/2|_(-2)^5 - 8t|_(-2)^5`

`d(t) = -5^3/3 + (-2)^3/3 + 3*25 - 3*4 - 8*5 - 8*2`

`d(t) = -133/3 +...

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