### Video Transcript

A vector-valued function or a
vector function is a function whose domain is a set of real numbers and whose range
is a set of vectors. We’re generally most interested in
vector functions 𝐫, whose values are three-dimensional vectors. In this video, we’re going to look
at how we can use our understanding of calculus to help us evaluate indefinite and
definite integrals of a vector function in both two or three dimensions. It’s therefore important you’re
confident in differentiating a range of functions, including polynomials,
trigonometric, and exponentials as well as the application of the various rules for
integration.

The definite integral of a
continuous vector function 𝐫 of 𝑡 can be defined in much the same way as for
real-valued functions except the integral is a vector. So we can say that the definite
integral between 𝑎 and 𝑏 of 𝐫 of 𝑡 with respect to 𝑡 is the limit as 𝑛
approaches ∞ of the sum of 𝐫 of 𝑡 𝑖 star times △𝑡 for values of 𝑖 from one to
𝑛. Then, we can express the integral
of 𝐫 in terms of the integrals of its component functions 𝑓, 𝑔, and ℎ as
shown. And then, using the definition of
the integral as being the limit of Riemann sums, we see that we can express the
definite integral between 𝑎 and 𝑏 of 𝐫 of 𝑡 with respect to 𝑡 as the definite
integral between 𝑎 and 𝑏 of 𝑓 of 𝑡 with respect to 𝑡 𝐢. Plus the definite integral between
𝑎 and 𝑏 of 𝑔 of 𝑡 with respect to 𝑡 𝐣. Plus the definite integral between
𝑎 and 𝑏 of ℎ of 𝑡 with respect to 𝑡 𝐤. And this is great because it means
we can simply evaluate an integral of a vector function by integrating each
component function.

We can even extend part of the
fundamental theorem of calculus to continuous vector functions such that the
definite integral between 𝑎 and 𝑏 of 𝐫 of 𝑡 with respect to 𝑡 is the
antiderivative capital 𝐑 evaluated at 𝑏 minus capital 𝐑 evaluated at 𝑎. So certainly, not rocket
science. We simply integrate each of our
components in the usual way. Let’s just see what that might look
like.

Calculate the indefinite integral
of four 𝑡 cubed plus three 𝑡 squared 𝐢 plus four 𝑡 squared minus five 𝐣 plus
four 𝑡 cubed minus five 𝑡 squared plus three 𝐤 with respect to 𝑡.

This is a vector-valued
function. This takes a real number 𝑡. And it outputs a position
vector. And to integrate a vector-valued
function, we simply integrate each component in the usual way. So we’ll integrate the component
for 𝐢 with respect to 𝑡. That’s four 𝑡 cubed plus three 𝑡
squared. We’ll integrate the component for
𝐣. That’s four 𝑡 squared minus
five. And we’ll integrate the component
for 𝐤, four 𝑡 cubed minus five 𝑡 squared plus three with respect to 𝑡. These are polynomial functions. And we know that, to integrate a
polynomial term whose exponent is not equal to negative one, we increase that
exponent by one and then divide by that number. This means the integral of four 𝑡
cubed is four 𝑡 to the fourth power over four. Integrating three 𝑡 squared and we
obtain three 𝑡 cubed over three. And of course, this is an
indefinite integral. So we must have that constant of
integration 𝑎. Simplifying fully and we obtain 𝑡
to the fourth power plus 𝑡 cubed plus 𝑎.

Similarly, when we integrate four
𝑡 squared, we get four 𝑡 cubed over three. The integral of negative five is
negative five 𝑡. And then, we need another constant
of integration 𝑏. In our final component, when we
integrate four 𝑡 cubed, once again, we get four 𝑡 to the fourth power over
four. The integral of negative five 𝑡
squared is negative five 𝑡 cubed over three. And the integral of three is three
𝑡. Let’s have the final constant of
integration 𝑐. This simplifies as shown.

We put this back into vector
form. And we see that our integral is
equal to 𝑡 to the fourth power plus 𝑡 cubed plus 𝑎 𝐢 plus four-thirds 𝑡 cubed
minus five 𝑡 plus 𝑏 𝐣 plus 𝑡 to the fourth power minus five-thirds 𝑡 cubed plus
three 𝑡 plus 𝑐 𝐤. Notice, though, that each of our
components has its own constant. So we can combine these and form a
constant vector 𝐜. And this means our integral is
equal to 𝑡 to the fourth power plus 𝑡 cubed 𝐢 plus four-thirds 𝑡 cubed minus
five 𝑡 𝐣 plus 𝑡 to the fourth power minus five-thirds 𝑡 cubed plus three 𝑡 𝐤
plus this capital 𝐂, which represents a constant vector.

We’ll now consider how this process
differs when evaluating a definite integral.

Evaluate the definite integral
between zero and 𝜋 by four of sec of 𝑡 tan of 𝑡 𝐢 plus 𝑡 times cos of two 𝑡 𝐣
plus sin squared of two 𝑡 times cos of two 𝑡 𝐤 with respect to 𝑡.

Remember, to integrate a
vector-valued function, we simply integrate each component in the usual way. Since this is a definite integral,
we can use the fundamental theorem of calculus, which says that the definite
integral between 𝑎 and 𝑏 of 𝐫 of 𝑡 with respect to 𝑡 is equal to the
antiderivative capital 𝐑 evaluated at 𝑏 minus capital 𝐑 evaluated at 𝑎. So essentially, all we’re going to
do is integrate each of our components with respect to 𝑡 and evaluate them
individually between the limits of zero and 𝜋 by four.

Let’s complete this process for the
component for 𝐢. It’s the definite integral between
zero and 𝜋 by four of sec 𝑡 tan 𝑡. We might recall the derivative of
sec of 𝑡 is equal to sec 𝑡 tan 𝑡. So this means the antiderivative of
sec 𝑡 tan 𝑡 must be sec 𝑡. So we use the fundamental theorem
of calculus. And we see that this is equal to
sec of 𝜋 by four minus sec of zero. But of course, sec of 𝑡 is equal
to one over cos of 𝑡. So we need to work out one over cos
of 𝜋 by four minus one over cos of zero. That’s root two minus one.

We’ve cleared some space so we can
evaluate the definite integral between zero and 𝜋 by four of 𝑡 times cos of two
𝑡. This time, we have the product of
two functions. So we’re going to use integration
by parts to evaluate. This says that the integral of 𝑢
times d𝑣 by d𝑥 with respect to 𝑥 is equal to 𝑢 times 𝑣 minus the integral of 𝑣
times d𝑢 by d𝑥 with respect to 𝑥. We’ll let 𝑢 be equal to 𝑡. And that’s because we know that d𝑢
by d𝑡 is simply one. This makes the second integral much
simpler. This means d𝑣 by d𝑡 must be cos
of two 𝑡. Then, the antiderivative of cos two
𝑡 gives us 𝑣. It’s a half sin of two 𝑡.

Now, of course, we’re working with
𝑡 as our parameter instead of 𝑥. But otherwise, the substitutions
are the same. We have 𝑢 times 𝑣. That’s 𝑡 times a half sin two 𝑡,
which we will shortly evaluate between zero and 𝜋 by four minus the definite
integral between those limits of a half sin two 𝑡 times d𝑢 by d𝑡, which is
one. We simplify this a little. And we see that we’re going to need
to work out the definite integral between zero and 𝜋 by four of a half of sin two
𝑡 with respect to 𝑡. Well, the antiderivative of a half
sin two 𝑡 is negative a quarter cos two 𝑡. And if we so wish, we can combine
these terms and then evaluate them between the limits of zero and 𝜋 by four. When we substitute 𝜋 by four, we
get 𝜋 over eight plus zero. And when we substitute zero, we get
zero plus a quarter. So that gives us 𝜋 by eight minus
a quarter.

Let’s clear some more space and
work out that final definite integral. This time, we’re going to use
integration by substitution. We’re going to let 𝑢 be equal to
sin two 𝑡. We know that d𝑢 by d𝑡, the
derivative of sin two 𝑡 with respect to 𝑡, is two cos two 𝑡. And whilst d𝑢 by d𝑡 isn’t a
fraction, we can equivalently rewrite this as a half d𝑢 equals cos of two 𝑡
d𝑡. We then need to change our
limit. So we use our substitution to do
so. Our lower limit is when 𝑡 is equal
to zero. And when 𝑡 is equal to zero, 𝑢 is
equal to sin of two times zero, which is zero. Then, when 𝑡 is equal to 𝜋 by
four, 𝑢 is equal to sin of two times 𝜋 by four, which is one.

We then replace sin two 𝑡 with
𝑢. We can replace cos of two 𝑡 with a
half d𝑢. And of course, we must make sure we
replace our limit. We might choose to take the
constant factor of a half outside our integral and focus on integrating 𝑢 squared
itself. That gives us 𝑢 cubed over
three. By substituting our limits, we get
a half times one cubed over three minus zero cubed over three, which is simply equal
to one-sixth. Now that we’ve evaluated the
integrals of each of our component functions, we’re ready to return this into vector
form. It’s root two minus one 𝐢 plus 𝜋
over eight minus a quarter 𝐣 plus one-sixth 𝐤.

Let’s have a look at another
example of the this form.

Evaluate the definite integral
between zero and one of the cube root of 𝑡 𝐢 plus one over 𝑡 plus one 𝐣 plus 𝑒
to the power of negative 𝑡 𝐤 with respect to 𝑡.

Remember, to integrate a
vector-valued function, we simply integrate each component in the usual way. And since this is a definite
integral, we can use the fundamental theorem of calculus and apply that to
continuous vector functions as shown. Here, capital 𝐑, of course, is the
antiderivative of lowercase 𝐫. So let’s integrate each component
with respect to 𝑡 and evaluate it between the limits of zero and one in the usual
way. We’ll begin by integrating the
component function for 𝐢. That’s the cube root of 𝑡 with
respect to 𝑡. And in fact, if we write that as 𝑡
to the power of one-third, it does become a little easier. We add one to the power and then
divide by that new value. So we get three-quarters 𝑡 to the
power of four over three. Then, evaluating this between the
limits of zero and one, we get three-quarters of one to the power of four over three
minus three-quarters of zero to the power of four over three, which is simply
three-quarters.

Next, we integrate the component
for 𝐣. That’s one over one plus 𝑡. We then recall that the indefinite
integral of 𝑓 prime of 𝑡 over 𝑓 of 𝑡 with respect to 𝑡 is equal to the natural
log of the absolute value of 𝑓 of 𝑡 plus some constant of integration 𝑐. Well, one is equal to the
derivative of one plus 𝑡. So the integral of one over one
plus 𝑡 is the natural log of the absolute value of one plus 𝑡. And we need to evaluate this
between zero and one. Notice how I’ve not included the
constant of integration here because this is a definite integral. And they cancel out. So we have the natural log of one
plus one minus the natural log of one plus zero. We no longer need the absolute
value sign because we know both one plus one and one plus zero are greater than
zero. So this is the natural log of two
minus the natural log of one, which is, of course, just the natural log of two.

Finally, we integrate our component
function for 𝐤. It’s the definite integral between
zero and one of 𝑒 to the power of negative 𝑡. The integral of 𝑒 to the power of
negative 𝑡 is negative 𝑒 to the power of negative 𝑡. So we have negative 𝑒 to the power
of negative one minus negative 𝑒 to the power of zero, which becomes negative 𝑒 to
the power of negative one plus one or one minus one over 𝑒. We’ve now successfully integrated
each of our component functions. So let’s put them back into vector
form. Our definite integral is equal to
three-quarters 𝐢 plus the natural log of two 𝐣 plus one minus one over 𝑒 𝐤. It’s important to realise that this
procedure can even be applied to initial value problems.

Solve the differential equation d𝐫
by d𝑡 equals 𝑡 squared plus three 𝐢 plus five 𝑡 𝐣 plus three 𝑡 squared 𝐤,
given that 𝐫 of zero equals five 𝐢 plus three 𝐣 minus two 𝐤.

Remember, we can begin by rewriting
our differential equation as d𝐫 equals 𝑡 squared plus three 𝐢 plus five 𝑡 𝐣
plus three 𝑡 squared 𝐤 d𝑡. And then, we can integrate both
sides of this equation. The integral of d𝐫 is the same as
the integral of one with respect to 𝐫. So it’s 𝐫 plus some constant of
integration 𝐜. Now, in fact, 𝐜 must be a vector
value since 𝐫 itself is a vector-valued function. We then integrate each component
function separately. And we get 𝑡 cubed over three plus
three 𝑡 plus some constant for the 𝐢, five 𝑡 squared over two plus some constant
for the 𝐣, and 𝑡 cubed plus some extra constant for the 𝐤. I’ve combined these into one
constant vector 𝐜 two.

Let’s subtract 𝐜 one from both
sides. And we see that we’re looking to
find the value of this constant vector 𝐜. We can use the fact that 𝐫 of zero
equals five 𝐢 plus three 𝐣 minus two 𝐤 to do so. So we substitute 𝐫 for five 𝐢
plus three 𝐣 minus two 𝐤 and 𝑡 for zero. And we get five 𝐢 plus three 𝐣
minus two 𝐤 equals zero 𝐢 plus zero 𝐣 plus zero 𝐤. Now, of course, we don’t need each
of these. And we see that 𝐜 must be equal to
five 𝐢 plus three 𝐣 minus two 𝐤. We can replace 𝐜 with this
vector. And then, we can combine the 𝐢s,
𝐣s, and 𝐤s. And when we do, we find that 𝐫 is
equal to 𝑡 cubed over three plus three 𝑡 plus five 𝐢 plus five 𝑡 squared over
two plus three 𝐣 plus 𝑡 cubed minus two 𝐤.

In this video, we’ve learned that
we can evaluate the integral of a vector-valued function by integrating each of its
component functions. This process can be applied to
definite and indefinite integrals. And we can even extend the
fundamental theorem of calculus to vector-valued functions such that the definite
integral between 𝑎 and 𝑏 of 𝐫 of 𝑡 with respect to 𝑡 is equal to the
antiderivative capital 𝐑 evaluated at 𝑏 minus the antiderivative evaluated at
𝑎.