Question

The functionbegin{cases}t & 0leq t<1 e^t & tgeq1 end{cases}has the following Laplace transform,L(f(t))=int_0^1te^{-st}dt+int_1^infty e^{-(s+1)t}dtTrue or False

Laplace transform
ANSWERED
asked 2021-01-13

The function
\(\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}\)
has the following Laplace transform,
\(L(f(t))=\int_0^1te^{-st}dt+\int_1^\infty e^{-(s+1)t}dt\)
True or False

Expert Answers (1)

2021-01-14

Step 1
Definition used -
Laplace transform of a function f(t) is given by -
\(F(s)=\int_0^\infty f(t)e^{-st}dt\)
Step 2
Given -
\(\begin{cases}t & 0\leq t<1\\ e^t & t\geq1 \end{cases}\)
\(L(f(t))=\int_0^\infty f(t)e^{-st}dt\)
\(=\int_0^1 te^{-st}dt+ \int_1^\infty e^te^{-st}dt\)
\(=\int_0^1 te^{-st}dt+ \int_1^\infty e^{(1-s)t}dt\)
Step 3
Answer - So given Laplace transform is False for function.

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