Transfer Function and Input Signal
The transfer function of a system describes the relationship between the input and output of the system in the frequency domain. In this case, the transfer function of the system is given as H(s) = 1/s, where s is the Laplace variable.

The input to the system is represented by x(t) = [sin(t) / ]u(t), where u(t) is the unit step function. This means that the input signal is a sinusoidal signal that starts at t = 0 and continues indefinitely.

Finding the Output of the System
To find the output of the system, we need to multiply the input signal by the transfer function and take the inverse Laplace transform.

Taking the Laplace transform of the input signal, we have:

X(s) = L{[sin(t) / ]u(t)}

Using the Laplace transform properties, we can rewrite the expression as:

X(s) = L{sin(t)} / s

The Laplace transform of sin(t) is given by:

L{sin(t)} = 1 / (s^2 + 1)

Substituting this into the expression for X(s), we get:

X(s) = (1 / (s^2 + 1)) / s

Simplifying further, we have:

X(s) = 1 / (s(s^2 + 1))

Now, we can multiply the transfer function H(s) = 1/s with X(s) to obtain the output Y(s) in the Laplace domain:

Y(s) = H(s) * X(s)
= (1/s) * (1 / (s(s^2 + 1)))
= 1 / (s^2(s^2 + 1))

Taking the Inverse Laplace Transform
To find the output y(t) in the time domain, we need to take the inverse Laplace transform of Y(s).

Using partial fraction decomposition, we can rewrite Y(s) as:

Y(s) = A/s + (B * s + C) / (s^2 + 1)

where A, B, and C are constants to be determined.

By equating the coefficients on both sides, we can solve for A, B, and C. After solving, we find that A = 0, B = 1, and C = 0.

Substituting these values back into the expression for Y(s), we have:

Y(s) = (s / (s^2 + 1))

Taking the inverse Laplace transform of Y(s), we obtain:

y(t) = L^(-1){(s / (s^2 + 1))}

Using the inverse Laplace transform table, we find that the inverse Laplace transform of s / (s^2 + 1) is given by:

L^(-1){(s / (s^2 + 1))} = cos(t)

Therefore, the output of the system is y(t) = cos(t).

Interpreting the Result
The correct answer is given as '0.45 to 0.55'. This means that the output of the system, y(t) = cos(t), lies between the values 0.45 and 0.55 for a certain range of t.

Since the