- [James] We're asked to find a Laplace transform,
which is a big f of s of the function f of t equals
e raised to the power of two t.
So applying the definition below,
we have the Laplace transform
of e raised to the power of two t equals big f of s
which equals the integral from zero to infinity
of f of t which is e raised to the power of two t times
e raised to the power of negative s-t d-t
and we're told that s is greater than two.
Notice in the integrated function we're multiplying
and the bases are the same so we can add the exponents.
We can write this as the integral from zero to infinity
of e raised to the power of two t minus s-t d-t
and now because we have an improper integral
because the upper limit of integration is infinity
we need to write this using limit notation.
We would have the limit, let's use the variable b,
so as b approaches infinity,
of the integral from zero,
because b is approaching infinity,
we can replace infinity with b.
And let's also write the exponent in factored form.
Let's factor out negative t and write this as
e raised to the power of negative t times the quantity
we'd have negative two plus s.
So let's write s minus two.
And now for our next step, to keep things organized,
let's find the indef. integral,
then we'll come back and find the def. integral,
then determine the limit.
So for our next step, we'll find the indef. integral
of e raised to the power of negative t times the quantity
s minus two integrated with respect to t.
We'll notice how we'll have to perform u-substitution,
where u is equal to negative t
times the quantity s minus two.
So differential u is equal to the derivative of this product
times d-t.
Well s minus two is just a constant.
The derivative of negative t with respect to t
is negative one so we have negative one times
the quantity s minus two
d-t.
So if we divide both sides by a negative one
times the quantity s minus two
we would have d-t equals, let's write this as
one over the opposite of the quantity s minus two d-u.
So in terms of u, we can write this as the integral,
this would be e to the u.
Again d-t is equal to one over
the opposite of the quantity s minus two d-u.
Which means the anti-derivative in terms of t
would be one over the opposite of the quantity s minus two
times e raised to the power of negative t
times the quantity s minus two plus c.
Remember our goal was to find
the def. integral then determine the limit.
So we can write this as the limit
as b approaches infinity
of one over the opposite of the quantity s minus two
times e raised to the power of negative t
times the quantity s minus two.
And the limits of integration are from t equals zero
to t equals b.
So now we'll find big f of b minus big f of a
then determine the limit.
Let's do this on the next slide.
We'll notice how this fraction here does not involve
b or t so we can write this as
one divided by the opposite of the quantity s minus two
times the limit as b approaches infinity of
e raised to the power of negative t
times the quantity s minus two.
Now we'll perform our substitutions for t.
So when t is equal to b,
you would have e raised to the power of negative b
times the quantity s minus two and then minus
when t is zero, we have e raised to the power of
negative zero times the quantity s minus two.
Well as b approaches infinity, this term approaches zero
and here we have e to the zero which equals one,
but to help us understand why this approaches zero,
if we wanted to, we can write this
as a fraction by moving the exponential term down
to the denominator.
So we could write this as
one divided by the opposite of the quantity s minus two
times the limit as b approaches infinity of
we can write this as, one over e raised to the power of
positive b times the quantity s minus two.
Again, this would be minus one.
Remember we're told that s is greater than two
so this quantity is positive,
and b is approaching infinity
so the exponent on e is approaching infinity,
which means the denominator of this fraction
is increasing without bound
and therefore this fraction is approaching zero.
So the limit is equal to negative one.
So we have one divided by
the opposite of the quantity s minus two times negative one.
Well we have a negative here and a negative one here,
so this product would just be one divided
by the quantity s minus two.
We just used Laplace transform
of e raised to the power of two t.
So to summarize:
the Laplace transform of e raised to the power of two t
equals big f of s which we now know equals one divided by
the quantity s minus two.
So going back to our first slide,
big f of s is equal to one divided by the quantity
s minus two.
If you're entering this from the keyboard,
it is important to put parentheses around
the quantity s minus two.
As I mentioned in the last video,
after we do a few of these,
we won't find the Laplace transform of a given function
using the definition, we use a table of Laplace transforms.
So let's see if we can find the formula in the table
that would allow us to find the Laplace transform
of this function much faster.
So here are just a few of the Laplace transform formulas.
So notice how if the function f of t equals e raised
to the power of a-t, then the Laplace transform of f of t,
or big f of s would be equal to one divided by the quantity
s minus a.
In our case, we had e raised to the power of two t,
so a was two, so we can quickly see that by applying
the definition we did find the Laplace transform correctly.
It would be one divided by the quantity s minus two,
where s is greater than two.
I hope you found this helpful.