Find Laplace Transformation f(t)=e^(t+7)

Scamuzzig2

Scamuzzig2

Answered question

2022-09-01

I worked out this far and I am having trouble with the integral
F ( s ) = 0 e s t f ( t ) d t
f ( t ) = e t + 7
0 e ( s 1 ) t e 7 d t
we can say that e 7 = c 1 and show
c 1 s 1 0 e ( s 1 ) t d t
I would surmise that I can take the limit at this point. So
c 1 s 1 lim b 0 b e ( s 1 ) t d t
which is likely evaluates as
c 1 s 1 ( 0 1 )
yielding
e 7 s 1
What is "good practice" that I have not shown?

Answer & Explanation

Haylie Campbell

Haylie Campbell

Beginner2022-09-02Added 13 answers

The only big problem is that when you've computed the antiderivative the integral becomes an evaluation:
c 1 s 1 lim u [ e ( s 1 ) t | 0 u ] = c 1 s 1 [ lim u 1 e ( s 1 ) u 1 e ( s 1 ) 0 ] = c 1 s 1 [ 0 1 ] = c 1 s 1 .
Domianpv

Domianpv

Beginner2022-09-03Added 2 answers

You have computed the antiderivative but kept the integral sign which is a mistake. Namely, we have
0 e ( s 1 ) t e 7 d t = e 7 ( lim M 0 M e ( s 1 ) t d t ) = e 7 ( lim M [ e ( s 1 ) t 1 s ] t = 0 t = M ) = e 7 ( lim M e ( s 1 ) M 1 1 s ) = e 7 s 1 .

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