Welcome to our lesson on definite
integration of vector valued functions.
To determine the definite integral of a vector
valued function, we integrate each component separately.
So looking at our first example,
because the vector value function only has
an x and y component, we'll have two definite integrals.
This is equal to the definite integral from zero to two
of e raised to the power of four t dt times i,
minus the definite integral from zero to two.
Instead of the square root, we'll use the 1/2 power.
So we'll have the quantity t plus one
to the 1/2 dt times j.
And now, we'll integrate.
Looking at this first integral,
notice how we'll have to use u substitution,
where u is equal to four t, and differential
u is equal to four dt.
So if we divide both sides by four,
provide this as e to the u, we'll have
an extra factor of 1/4, which means
the anti-derivative is 1/4 e to the four t.
But remember, this is definite integration,
so we have our limits of integration from zero to two
times i and then minus.
This may appear it requires u substitution,
but notice the inner function is u equals t plus one,
therefore differential u is dt.
So we can use the basic power rule here.
So we'll add one to the exponent
and then divide by that new exponent.
1/2 plus one is 3/2.
And then instead of dividing by 3/2,
we'll multiply by the reciprocal.
So we'll have 2/3 times the quantity
t plus one to the 3/2 power.
Again, zero to two times j.
Now, we'll evaluate.
We'll first substitute two for t.
So we'll have 1/4 e to the 8th.
Four times two is eight,
minus 1/4 e to the zero times i.
And then here, we'll have 2/3
when t is two we'll have times three to the 3/2,
and then minus 2/3 times one to the 3/2, times j.
Let's go ahead and factor out 1/4 here.
E to the zero is equal to one.
So we'll have 1/4 times e to the 8th
minus one times i, and then minus.
Here, we'll factor out 2/3,
leaving us with three to the 3/2 minus one times j.
Now, let's take a look at a second example.
Notice how the vector valued function
has an x, y, and z component.
So now, we'll have three definite integrals.
Now when we integrate, notice how here,
we have to perform u substitution,
where we'll let u equal sine t,
therefore differentially, u is equal to cosine tdt,
which is a perfect fit.
We can write this in terms of u as the integral
of u squared, and because cosine tdt is equal to du,
this is just du, which means
the anti-derivative in terms of t
would just be u to the 3rd divided by three,
or sine cubed t divided by three.
And now for the next integral, notice that u
is equal to cosine t this time,
because this is raised to the second power.
So differential u is equal to negative sign tdt,
which means negative du is equal to sine tdt.
So in terms of u, this would be negative u squared du.
Again, sine tdt is equal to negative du,
giving us this negative.
And then for the anti-derivative in terms of u,
we'll have negative u to the third divided by three,
which would be in terms of t,
minus cosine cubed t divided by three.
And then finally, the integral of two
with respect to t would be 2t.
So we have 2t, times k.
So we first substitute t equals pi over two.
Sine pi over two is one,
so we'll have one cubed divided by three,
that's 1/3 minus.
When t is zero, sine zero is zero,
so it would be minus zero, times I,
minus cosine of pi over two is zero.
So we'd have zero minus consine zero is one.
So we'd have one cubed divided by three or 1/3,
times j, and then, plus.
Notice when t is pi over two, we have
two times pi over two, which is just pi.
And then when t is zero, it's minus zero times k.
So, we're left with 1/3 i.
This is minus negative 1/3 j or plus 1/3 j.
And then, plus pi k.
Okay, I hope you found this explanation helpful.