Integral Fun?! A particle moves along a straight line with velocity v⁡t=t+2 meters/second.?

Question

A particle moves along a straight line with velocity v⁡(t)=t+2 meters/second. Assume that the time t is measured in seconds. Suppose that t[0] .. t[3] divide the interval [2, 8] into 3 equal intervals. Use these subintervals and the right endpoints as sample points to compute a Riemann sum which approximates the distance traveled by the particle for the interval of time [2, 8].

Answer 1

Using right endpoints, the sum is

v(t[1])(t[1] - t[0]) + v(t[2])(t[2] - t[1]) + v(t[3])(t[3] - t[2])

Since the intervals are equal, call it Δt, the sum is

Δt(v(t[1]) + v(t[2]) + v(t[3]))

which here is

2(v(4)) +v(6) + v(8))

You should get 48

Answer 2

to get displacement, we ought to combine: ds/dt = t^2 e^{-3t} utilising integration by utilising factors: (crucial of) u dv/dt dt= uv - {(crucial of) v du/dt dt} u = t^2 dv/dt = e^{-3t} du/dt = 2t v = -(a million/3) (e^{-3t}) s = -(a million/3)(t^2)(e^{-3t}) - {(crucial of) -(a million/3) (e^{-3t}) 2t dt } utilising integration by utilising factors back on the 2d element of that provides: s= -(a million/3)(t^2)(e^{-3t}) + a million/3((-2t/3)(e^{-3t}) -{(crucial of) (-a million/3)e^{-3t} dt } ) which simplifies to: s= -(a million/27)(e^{-3t})(9t^2 +6t + 2) + c for t = 0, we ought to constantly have s=0 on account that's the place we are taking this relative from, so c = 2/27. so the respond is s= 2/27 - (a million/27(e^{-3t})(9t^2 + 6t + 2)) notice- quasar's exceeded over the mixing consistent, in any different case it rather is an equivalent expression, mr phatt's used accepted speed- if speed relies upon on time, you need to use integration.