You can do it the easy way, or the hard way

Calculus Level pending

$F(s) = \frac{2}{(s+1)^2+4}$

Given that $F(s)$ is the Laplace transform of $f(t)$ , what is the area underneath the curve $y = f(t)$ on the interval $[0, \infty)$ ? Try to do it without using an inverse Laplace transform.

Hint: The defenition of the Laplace transform (notated as $\mathcal{L}(f(t))$ or $F(s)$ ) is:

$\int_0 ^ {\infty} f(t)e^{-st}dt$

The answer is 0.4.

2 solutions

Chew-Seong Cheong
Jun 12, 2020

$\begin{aligned} F(s) & = \int_0^\infty f(t) e^{-st} dt = \frac 2{(s+1)^2+4} \\ \implies F(0) & = \int_0^\infty f(t) dt = \frac 2{(0+1)^2+4} = \frac 25 = \boxed{0.4} & \small \blue{\text{The area under }f(x) \in [0, \infty)} \end{aligned}$

Ron Gallagher
Jun 12, 2020

Note that F(0) is the integral from 0 to infinity of f, which is the required area. F(0) = 2/5 = .4

While the "shortcut" method I used is easier that taking the inverse Laplace Transform and computing the integral, we should be careful. Implicit in this method is the assumption that the integral from 0 to infinity of f exists. There are functions for which the Laplace Transform exists, but the integral from 0 to infinity is divergent (for example, g(x) = sin(x) is one such function).

Ron Gallagher - 12 months ago

You are absolutely right. More deterministic method is to use inverse Laplace transform. The inverse transform of the given expression is $e^{-t}\sin (2t)$ , which, when integrated within the given limits, yields $\dfrac{2}{5}=0.4$ .

A Former Brilliant Member - 12 months ago

You can do it the easy way, or the hard way

The answer is 0.4.

2 solutions

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