Video: Integrals of Vector-Valued Functions

In this video, we will learn how to integrate vector-valued functions in 2D and 3D.

13:16

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.

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