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F (s) is the Laplace transform of the function
f (t) = 2t2e-t
F(1) is __________ (correct to two decimal places).
    Correct answer is '(0.5)'. Can you explain this answer?
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    Laplace Transform and its Definition:

    The Laplace transform is a mathematical operation that transforms a function of a real variable t into a function of a complex variable s. It is commonly used in engineering and physics to solve differential equations and analyze systems in the frequency domain.

    The Laplace transform of a function f(t), denoted as F(s), is defined as:

    F(s) = L{f(t)} = ∫[0 to ∞] e^(-st) f(t) dt

    where s is a complex variable and t is the time variable.

    Given Function and its Laplace Transform:

    In this problem, we are given the function f(t) = 2t^2e^(-t) and we need to find its Laplace transform F(s).

    To find the Laplace transform, we substitute the given function into the Laplace transform definition:

    F(s) = L{f(t)} = ∫[0 to ∞] e^(-st) (2t^2e^(-t)) dt

    Simplifying the expression inside the integral:

    F(s) = 2 ∫[0 to ∞] t^2 e^(-st) e^(-t) dt
    = 2 ∫[0 to ∞] t^2 e^(-(s+1)t) dt

    Integration by Parts:

    To solve this integral, we can use the technique of integration by parts. The formula for integration by parts is:

    ∫u dv = uv - ∫v du

    We choose u = t^2 and dv = e^(-(s+1)t) dt. Then, we can find du and v by differentiating and integrating u and dv, respectively.

    du = 2t dt
    v = -1/(s+1) e^(-(s+1)t)

    Substituting these values into the integration by parts formula:

    ∫ t^2 e^(-(s+1)t) dt = (-t^2/(s+1)) e^(-(s+1)t) - ∫ (-2t/(s+1)) e^(-(s+1)t) dt

    Recursive Integration by Parts:

    We can see that we have a similar integral on the right-hand side of the equation. We can apply integration by parts again to solve it.

    Let's denote the second integral as I. We choose u = -2t/(s+1) and dv = e^(-(s+1)t) dt. Then, we can find du and v by differentiating and integrating u and dv, respectively.

    du = -2/(s+1) dt
    v = -1/(s+1) e^(-(s+1)t)

    Substituting these values into the integration by parts formula:

    I = (-2t/(s+1)) e^(-(s+1)t) - ∫ (-2/(s+1)) (-1/(s+1)) e^(-(s+1)t) dt
    = (-2t/(s+1)) e^(-(s+1)t) + (2/(s+1))^2 e^(-(s+1)t) - ∫ (2/(s+1))^2 e^(-(s+1)t) dt

    We can observe that we have another similar integral on the right-hand side.
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