Question

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the improper integral L{f(t)}(s) = infinity 0 e-st.f(t)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of as a fixed constant)

1. Find L{t}. (hint: remember integration by parts)
A. 1
B. -1/s2
C. 0
D. 1/s2
E. -s2
F. None of these
2. Find L{1}.
a.1/s
b. 1
c. 0
d. -s
e. -1/s
f. none of these

Answer 1

(1) D

`L_s\left\{t\right\} = \displaystyle\int_0^\infty te^(-st)\,\mathrm dt`

Integrate by parts, taking

`u = t \implies \mathrm du=\mathrm dt`

`\mathrm dv = e^(-st)\,\mathrm dt \implies v=-\frac1se^(-st)`

Then

`L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^(-st)\right]\bigg|_(t=0)^(t\to\infty)+\frac1s\int_0^\infty e^(-st)\,\mathrm dt`

`L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^(-st)\,\mathrm dt`

`L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^(-st)\bigg|_(t=0)^(t\to\infty)`

`L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}`

(2) A

`L_s\left\{1\right\} = \displaystyle\int_0^\infty e^(-st)\,\mathrm dt`

`L_s\left\{1\right\} = \displaystyle\left[-\frac1se^(-st)\right]\bigg|_(t=0)^(t\to\infty)`

`L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}`