Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).

So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.