Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the Laplace transform of f, provided that the integral converges. to find ℒ{f(t)}. (Write your answer as a function of s.) f(t) = cos(t), 0 ≤ t < π 0, t ≥ π

L(f(t)) =

Step-by-step explanation:

let f be a function defined for t ≥ 0

we can write the function f(t) in terms of unit function as follows

f(t) = 2 u,(t) - 1  where

      0≤ t < 1

f(t) = (2 * 0) -1 = -1

when t ≥ 1

f(t) = (2*1 )- 1 = 1

Now the Laplace transform L(F(T)) = 2L( u, (t) ) - L(1)  --------equation 1

this is because L(u,(t)) =

c = 1 hence L(1) = 1/s

back to equation 1

L(f(t)) = 2 -  1/s  laplace transform

also L(u(t) ) =

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.

,jkulgiurutrjyftkugyi.uykujuyxjdkytuf,yukj

step-by-step explanation: