### Video Transcript

Calculate the integral of π‘π’ plus three π£ with respect to π‘.

Weβre given an indefinite integral which weβre asked to evaluate. And we can see something interesting about our integrand, In this case, weβre given a vector-valued function. And we know this is a vector-valued function because it contains the unit directional vectors π’ and π£. And itβs also worth pointing out here thereβs 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 doesnβt 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, weβll start with evaluating the integral of our first component function. Thatβs the integral of π‘ with respect to π‘. And we can evaluate this integral by using the power rule for integration. We recall this tells us for all constants π not equal to negative one, the integral of π‘ to the πth power with respect to π‘ is equal to π‘ to the power of π plus one divided by π plus one plus the constant of integration π.

In our case, we can write π‘ as π‘ to the first power. So, our value of π 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 π‘ squared over two. And weβll add a constant of integration weβll call π. Next, we need to evaluate the integral of our second component function. Thatβs the integral of three with respect to π‘.

Thereβs a few different ways of doing this. For example, we could rewrite three as three times π‘ to the zeroth power and then use the power rule for integration. If we did this, we would get three π‘ to the first power divided by one plus the constant of integration π. But of course, dividing by one and raising a number to the first power doesnβt change this value. So, we just get three π‘ plus π.

Weβre now ready to use these to evaluate the integral of our vector-valued function. That is the integral of π‘π’ plus three π£ with respect to π‘. Now, all we need to do is integrate this component wise. And we found the integral of our first component is π‘ squared over two plus π and the integral of our second component is three π‘ plus π. So, we get the integral of our vector-valued function with respect to π‘ is π‘ squared over two plus π times π’ plus three π‘ plus π times π£. And we could leave our answer like this. However, thereβs one more piece of simplification we could do.

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

Therefore, by evaluating the integral of the vector-valued function π‘π’ plus three π£ with respect to π‘ componentwise, we were able to show that this must be equal to π‘ squared over two π’ plus three π‘π£ plus π.