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The velocity function is v(t) = t^2 - 6 t + 8 for a particle moving along a line. Find the displacement and the distance traveled by the particle during the time interval [-1,6].

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    v=ds/dt

    ds/dt=t^2-6t+8

    ds=(t^2-6t+8)*dt

    s=integral(t^2-6t+8)*dt

    s=(t^3-9t^2+24t)/3+C

    s=(t/3)*(t^2-9t+24)+C

    s=(t/3)*[(t-9)*t+24)]+C

    s=Definite integral betwen(-1,6)

    s=(6/3)*[(6-9)*6+24]-(-1/3)*[(-1-9)*(-1)+24]

    s=(6/3)*[(-3)*6+24]+(1/3)*[-10*(-1)+24]

    s=2*(-18+24)+(1/3)*(10+24)

    s=2*6+(1/3)*34
    s=12+34/3

    s=(36/3)+(34/3)

    s=70/3

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