F(s) is the Laplace transform of the function f(t) = 2t²e

F(s) is the Laplace transform of the function f(t) = 2t²e^–t
F(1) is _______(correct to two decimal places.)

Applying Laplace transform

L[teⁿ] =	n!
	Sⁿ⁺¹

L[t²] =	2
	S³

L[2t²] =	4
	S³

F(S) = L[e^-t.2t²] =	4
	(S + 1)³

Putting s = 1 we get

F(1) =	4	=	4	= 0.5
	(1 + 1)³		8