Use Definition 7.1.1. DEFINITION 7.1.1 Laplace Transform Let...

Use Definition 7.1.1. DEFINITION 7.1.1 Laplace Transform Let f be a function defined for t ? 0. Then the integral ?{f(t)} = ??? e^-st f(t) dt is said to be the Laplace transform of f, provided that the integral converges. Find ?{f(t)}. (Write your answer as a function of s.) f(t) = {t, 0 ? t < 1 {1, t ? 1 ?{f(t)} = -1/s² (e^-s) + 1/s² (s > 0)

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00:01 Hi here for the given question we are given that here we have f of s is equal to integration over 0 to infinity e to the power minus st f of t dt.

00:13 So here in our case our function is here we have f of t equals to integration like t for 0 less than or equal to t less than 1 and here in our case another value is 1 for t greater than or equal to 1.

00:39 So here in our case now we can say that f of s can be written as integration over 0 to 1 e to the power minus st into t dt.

00:49 This is for the interval 0 less than or equal to t less than 1 and f of s equals to integration over 1 to infinity for e to the power minus st dt.

01:00 So here this is for t greater than or equal to 1.

01:03 So here we will tell this as first part and this will be our second part.

01:08 So here for the first part we know that here in our case if we use u substitution method then integration over 0 to 1 e to the power minus st into t dt can be further written as here we have value equals to minus t by s e to the power minus st minus minus will be plus 1 upon s e to the power minus st dt.

01:35 So here now further on integrating this term and simplifying further we have minus t by s e to the power minus st plus 1 upon s square e to the power minus st and here the limit of integration are from 0 to 1 here also 0 to 1.

01:50 So here now substituting the value and solving further we can say that here we have this value equals to minus 1 upon s e to the power minus s plus 1 upon s square e to the power minus s minus 1 upon s square.

02:08 So here in our case now this is the value of f of t for the first part.

02:14 Now similarly we need to calculate for the second part.

02:17 So here in our case for the second part we have integration over 1 to infinity e to the power minus st dt...

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