We're at the final Laplace Transform
which is big F of s of the function f of t equals t cubed.
So looking at the definition of a Laplace Transform
here at the bottom,
we have a Laplace Transform
of t to the third is equal to big F of s
which equals the integral from zero to infinity
of f of t which is t cubed times e
raised to the power of negative s t, d t.
Notice how here we have an improper integral
so we need to rewrite this using limit notation.
This is equal to the limit as b approaches infinity
of the integral from, instead of from zero to infinity,
we'd have from zero to b, again because b is
approaching infinity, of t to the third times e raised
to the power of negative s t, d t.
Now for our next step, we'll find the indef integral
of t to the third, e to the negative s t, with respect to t.
Then we'll come back here and find the def integral
and then determine the limit.
So our next step,
let's find the integral of t to the third,
e to the negative s t with respect to t.
This is going to take quite a bit of work
because we'll have to apply integration by parts,
shown here below, multiple times.
To begin, we'll let u equal t to the third
which means differential v is equal to e raised to the power
of negative s t, d t.
So we differentiate to find differential u
and we integrate to find v.
The derivative of t to the third with respect to t
would be three t squared.
So differential u equals three t squared d t.
And now I need to integrate both sides
of the equation here to find v.
So we'd integrate e to the negative s t with respect to t.
This requires u substitution where u equals negative s t.
Differential u equals negative s d t,
so dividing both sides by negative s,
we have negative one over s,
differential u equals d t.
So we integrate here, we'll have an extra factor
of negative one divided by s.
V is equal to negative one divided by s
e to the negative s t.
So applying the integration by parts formula,
we have u times v which should be negative one
divided by s, t to the third, e to the negative s t
minus integral of v d u.
Looking at v times d u,
notice how that product will be negative.
Because we have a minus here it becomes plus.
Then let's write this as three divided by s.
The integrated function would be t to the second
e to the negative s t, d t.
And now to integrate here we'll have
to perform u substitution again.
So to integrate here it will be similar to what we did here
except now u equals t squared,
so differential u equals two t d t,
but differential v is still e raised
to the power of negative s t, d t.
So when we integrate here to find v,
we get v equals negative one divided by s
e to the negative s t.
Need to be careful here though because we're going
to have three divided by s times the quantity u v
minus the integral of v d u.
So this first term stays the same,
we have negative one divided by s
t to the third, e to the negative s t
and then we'll have plus three divided by s
times u times v is going to be negative one divided by s
t to the second, e to the negative s t
minus the integral of v d u.
Again we have minus here.
V times the u is negative so this becomes plus.
Let's write this as two divided by s times the integral of t
e to the negative s t, d t.
Before we integrate here though, let's distribute
the three divided by s to keep things more organized.
So the first term stays the same.
Then we'll have minus three divided by s squared, t squared
e to the negative s t.
Then we'll have plus six divided by s squared
times the integral of t, e to the negative s t, d t.
And now I'll perform integration by parts again here
where we'd have u equals t so differential u equals d t.
Differential v is equal to e raised
to the power of negative s t, d t.
So integrating here to find v, we get v equals again,
negative one divided by s, e to the negative s t.
So these first two terms stay the same
and we're going to have six divided by s squared
times the quantity u times v.
Well, u times v is negative one divided by s, t,
e to the negative s t and then minus
the integral of v d u, again v times d u is negative
because we have minus here becomes plus.
So we have plus one over s times the integral
of e raised to the power of negative s t, d t.
Again, to keep things organized,
let's distribute six divided by s squared here and here.
So the first two terms stay the same and then we have minus
six divided by s cubed t, e to the negative s t
plus six divided by s cubed times the integral of e
raised to the power of negative s t, d t.
Now to integrate e raised to the power of negative s t, d t,
before I've done that several times up here,
so the first three terms stay the same.
So we have plus six divided by s cubed
times the integral of e raised to the power of negative s t
with respect to t is negative one divided by s,
e to the negative s t.
And of course we have plus that constant of integration.
Let's go ahead and find this product
and write this one more time on the next slide.
So for this last step to find the indef integral,
let's change the form of these first three terms
and then we'll find this product.
So let's write this first term as negative t to the third
divided by s, e raised to the power of negative s t.
Let's write the second term as minus three t squared
divided by s squared, e to the negative s t.
Let's write the third term as minus six t divided by s cubed
e to the negative s t.
And let's write this product as minus six divided
by s to the fourth, e to the negative s t
and then plus C.
So this is the indef integral.
Remember our goal was to find this definite integral
and determine the limit.
So now that we know the anti-derivative
or the indef integral though, we can write this as
the limit as b approaches infinity of negative t
to the third divided by s, e to the negative s t
minus three t squared divided by s squared,
e to the negative s t minus six t divided by s cubed
e to the negative s t and then finally,
minus six divided by s to the fourth, e to the negative s t.
And again the limits of integration are
from t equals zero to t equals b.
And now we need to find big F of b minus big F of a
so I'll first substitute b for t.
So I have the limit as b approaches infinity of,
again we're substituting b for t,
so we'll have negative b to the third divided by s
times e raised to the power of negative s b
minus three b squared divided by s squared
times e raised to the power of negative s b
minus six b divided by s to the third
times e raised to the power of negative s b
and finally minus six divided by s to the fourth
times e raised to the power of negative s b
and now we substitute zero for t.
Notice all these terms contain a factor of t
except this last term, so these first three terms
would be zero because we'd have zero cubed here,
zero squared here, zero here.
So we're just left with minus six divided by s to the fourth
times e raised to the power of, this'd be negative s
times zero which would be e to the zero.
And now for the terminus limit on the next slide.
Now to determine the limit as b approaches infinity,
if we needed to, we could write these first four terms
by moving the exponential term down to the denominator.
For example, in this first term we could write this
as negative b raised to the power of three
divided by s times e raised to the power of positive s b.
To determine the limit as b approaches infinity in this form
we may need to apply L'Hopital's Rule several times,
but the limit will be equal to zero.
So these four terms all approach zero
as b approaches infinity, so we're just left
with minus negative six divided by s to the fourth
times e to the zero, which is one.
So this limit is equal to negative and then negative six
divided by s to the fourth, which gives us positive six
divided by s to the fourth which is the Laplace Transform
of f of t equals two to the third.
So to summarize, the Laplace Transform
of t to the third which equals big F of s is
equal to six divided by s to the fourth.
So going back to our first slide, again we now know
big F of s is equal to six divided by s to the fourth.
Now this was quite a bit of work
and just like there are formulas for derivatives,
there are formulas for Laplace Transforms.
So it would've been much faster if we use
the Laplace Transform formula shown here
where if we have f of t equals t to the third,
then n equals three, which means the Laplace Transform,
big F of s, is equal to three factorial
divided by s raised to the power of three plus one,
or s to the fourth.
So again, knowing that f of t is equal to t to the third,
we can quickly determine the Laplace Transform
of t to the third which equals big F of s is equal to,
we have t to the third, so n is three,
so we'd have three factorial divided by s
raised to the power of three plus one.
Well, three factorial is three times two times one
divided by s to the fourth, which of course
gives us six divided by s to the fourth.
So after we do several of these using the definition
of a Laplace Transform, we'll then just use the formulas
to determine the Laplace Transforms.
I hope you found this helpful.