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) ) =

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )

step-by-step explanation:

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