Find the inverse Laplace transform of the following function using the theorem of transforms of integrals. \[ F(s)=\frac{1}{s\left(s^{2}+16\right)} \] The solution is $\mathrm{f}(\mathrm{t})=$ (Type an expression using $t$ as the variable. Type an exact answer.)

Step 1 ：The given function is a product of two functions, \(\frac{1}{s}\) and \(\frac{1}{s^{2} + 16}\).

Step 2 ：The inverse Laplace transform of \(\frac{1}{s}\) is 1 and the inverse Laplace transform of \(\frac{1}{s^{2} + 16}\) is \(\sin(4t)\).

Step 3 ：The theorem of transforms of integrals states that the Laplace transform of the integral of a function is equal to the original function divided by s.

Step 4 ：So, to find the inverse Laplace transform of the given function, we need to find the inverse Laplace transform of each part and then multiply them together.

Step 5 ：The inverse Laplace transform of the given function \(F(s)\) is \(-\cos(4t)\cdot\text{Heaviside}(t)/16 + \text{Heaviside}(t)/16\). The Heaviside function is the unit step function, which is 0 for \(t < 0\) and 1 for \(t \geq 0\).

Step 6 ：This means that the inverse Laplace transform is 0 for \(t < 0\) and \(-\cos(4t)/16 + 1/16\) for \(t \geq 0\).

Step 7 ：Final Answer: For \(t < 0\), \(f(t) = 0\). For \(t \geq 0\), \(f(t) = \boxed{-\frac{\cos(4t)}{16} + \frac{1}{16}}\).