Verified answer

Velocity is given in distance/time, which is basically dx/dt (where x = displacement). In order to find the displacement, you need to take the integral of the velocity equation on [0, 4]:

∫ v(t) dt

∫ 10t / (1+5t^2) dt

Let:

u = 1 + 5t^2

du = 10t dt

When t=0, u=1

When t=4, u=81

The integral becomes:

∫ 1/u du

ln|u|

from 1 to 81:

ln81 - ln1

ln81 or approximately 4.394

If s(0) = 3, you just have to add 3 to the displacement to find the final position:

ln81 + 3

or

4.394 + 3 = 7.394

Displacement = s(t) = ∫ v(t) = ∫ 10t/ (1+5t^2) dt , 0≤t≤4

∫ 10t/ (1+5t^2) dt

Let u=1+5t^2

du = 10t dt

∫ 10t/ (1+5t^2) dt = ∫ du/u = ln(u)

= ln(1+5t^2)

s(4) = ln(81)

s(0) = 3

At t=4, s(t)= ln(81)+3

= 7.3944

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