Welcome to a lesson
on determining the antiderivative
of a vectored valued function.
This is also called Indefinite Integration
of a vectored valued function.
Notice, to perform indefinite integration
of a vectored valued function, we integrate each component
of the vectored valued function.
It's important to remember
that the antiderivative of a vectored valued function
is a family of vectored valued functions,
such that if big R of t is the antiderivative,
then the derivative of big R of t,
is equal to little r of t.
And in some cases, we are given an initial condition,
and in this case, we can find the
exact antiderivative vectored valued function.
We'll take a look at this case in the next video.
Let's take a look at our first example.
And again, in order to find this indefinite integral,
we're going to integrate each component,
which means, we'll have the indefinite integral of
five divided by t squared, which we'll write as
five, t to the power of negative two, dt,
times i, and then, we'll have minus
the integral of four, square root of t, which we'll write as
four, t to the power of 1/2, dt,
times j.
And now, we'll apply the Power Rule of integration
to find our indefinite integrals.
So, we're going to add one to the exponent,
and then, divide by the new exponent,
so we'll have five times, t to the power of negative two,
plus one is negative one,
divided by negative one, plus our constant of integration,
which we'll call c sub one, times i, minus
four times, t to the power of 1/2, plus one,
that's t to the 3/2,
divided by 3/2, which is the same as
multiplying by the reciprocal, or multiplying by 2/3,
plus c sub two, times j.
Now, let's go ahead and clean this up.
Notice how we have five divided by negative one,
that's negative five, t to the negative one,
or negative five divided by t, plus c sub one,
times i,
minus, here, nothing simplifies, so we have 8/3,
t to the 3/2,
plus c sub two, times j,
so this is the antiderivative,
but notice how c sub one, and c sub two,
result in a constant vector,
where the i component is c sub one,
and the j component would be negative c sub two,
because of the minus.
So, this can be written a different way.
If we let vector c, our constant vector
be equal to c sub one, i,
then it would be minus c sub two, j.
We can give the antiderivative as
negative five, divided by t, i,
minus 8/3, t to the 3/2,
plus our constant vector c.
Let's take a look at a second example.
Notice in this case,
our vectored valued function has three components,
so now, we'll have three indefinite integrals.
Let's write the first indefinite integral as
two, times the integral of one divided by t, dt,
times i, minus
the integral of sine t, dt, j,
and then, we'll have plus
the integral of secant squared, two t, dt, k.
Well, the integral of one divided by t, with respect to t,
would be natural log t, so here we have two, natural log t,
plus c sub one, i.
The integral of sine t is negative cosine t,
so because of this minus, we'll have plus
cosine t, plus c sub two, j,
and now, for this last integral,
we'll have to perform U substitution,
where the inner function, two t will be u,
so if u is equal to two t,
differential u is equal to two dt.
Dividing both sides by two,
notice how dt is equal to 1/2 du,
so if we wrote this in terms of u, we would have
1/2, the integral of secant squared, u, du.
And since the integral of secant squared u
is equal to tangent u, we'll have plus 1/2 tangent,
instead of u, two t,
plus c sub three, times k.
So now, we do have our antiderivative,
or the indefinite integral,
but because of the c sub one, c sub two, and c sub three,
we can let vector c, our constant vector,
be equal to c sub one, i,
plus c sub two, j,
plus c sub three, k.
So now, we can write this as
two, natural log t, i,
plus cosine t, j,
plus 1/2, tangent two t, k,
plus our constant vector, c.
That'll do it for this lesson.
In the next lesson, we'll talk about
how we can do these types of problems
when we're given an initial condition,
and therefore, find the exact antiderivative function.
I hope you found this helpful.