Video: Deteremining the Indefinite Integral of a Vector-Valued Function

Calculate ∫(𝑑𝐒 + 3𝐣)d𝑑.

03:02

Video Transcript

Calculate the integral of 𝑑𝐒 plus three 𝐣 with respect to 𝑑.

We’re given an indefinite integral which we’re asked to evaluate. And we can see something interesting about our integrand, In this case, we’re given a vector-valued function. And we know this is a vector-valued function because it contains the unit directional vectors 𝐒 and 𝐣. And it’s also worth pointing out here there’s a lot of different notation for vectors. For example, we can see the hat notation, which represents unit vectors. However, you often also see vectors written in bold notation, underlined with full arrows or half-arrows. It doesn’t matter which notation you prefer; they all mean the same thing.

So now, we need to recall how to integrate a vector-valued function. To do this, we just need to evaluate the integral of each of our component functions. So, we’ll start with evaluating the integral of our first component function. That’s the integral of 𝑑 with respect to 𝑑. And we can evaluate this integral by using the power rule for integration. We recall this tells us for all constants 𝑛 not equal to negative one, the integral of 𝑑 to the 𝑛th power with respect to 𝑑 is equal to 𝑑 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝘊.

In our case, we can write 𝑑 as 𝑑 to the first power. So, our value of 𝑛 is equal to one. So, we add one to our exponent of one, giving us two, and then divide by this new exponent. This gives us 𝑑 squared over two. And we’ll add a constant of integration we’ll call π‘Ž. Next, we need to evaluate the integral of our second component function. That’s the integral of three with respect to 𝑑.

There’s a few different ways of doing this. For example, we could rewrite three as three times 𝑑 to the zeroth power and then use the power rule for integration. If we did this, we would get three 𝑑 to the first power divided by one plus the constant of integration 𝑏. But of course, dividing by one and raising a number to the first power doesn’t change this value. So, we just get three 𝑑 plus 𝑏.

We’re now ready to use these to evaluate the integral of our vector-valued function. That is the integral of 𝑑𝐒 plus three 𝐣 with respect to 𝑑. Now, all we need to do is integrate this component wise. And we found the integral of our first component is 𝑑 squared over two plus π‘Ž and the integral of our second component is three 𝑑 plus 𝑏. So, we get the integral of our vector-valued function with respect to 𝑑 is 𝑑 squared over two plus π‘Ž times 𝐒 plus three 𝑑 plus 𝑏 times 𝐣. And we could leave our answer like this. However, there’s one more piece of simplification we could do.

If we were to distribute our parentheses, we see we would end up with π‘Ž times 𝐒 plus 𝑏 times 𝐣. But remember, π‘Ž and 𝑏 are constants of integration. So, this is just equal to a constant vector. So, we can just call this constant vector the vector 𝐜. This gives us 𝑑 squared over two 𝐒 plus three 𝑑𝐣 plus our constant vector 𝐜. And this is our final answer.

Therefore, by evaluating the integral of the vector-valued function 𝑑𝐒 plus three 𝐣 with respect to 𝑑 componentwise, we were able to show that this must be equal to 𝑑 squared over two 𝐒 plus three 𝑑𝐣 plus 𝐜.

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