It looks like the particle's acceleration is
a(t) = 3t²
and we're given the initial velocity and position to be v(1) = 2 and x(1) = 6.
(a) By the fundamental theorem of calculus, the velocity at time t is
v(t) = v(1) + ∫₁ᵗ a(u) du
v(t) = 2 + ∫₁ᵗ 3u² du
v(t) = 2 + u³ |₁ᵗ
v(t) = 2 + (t³ - 1³)
v(t) = t³ + 1
(b) Use the theorem again to get the position at time t,
x(t) = x(1) + ∫₁ᵗ v(u) du
x(t) = 6 + ∫₁ᵗ (u³ + 1) du
x(t) = 6 + (1/4 u⁴ + u) |₁ᵗ
x(t) = 6 + (1/4 t⁴ + t - 1/4 • 1⁴ - 1)
x(t) = 1/4 t⁴ + t + 19/4
(c) The position of the particle at t = e is
x(e) = 1/4 e⁴ + e + 19/4