### Video Transcript

Calculate the integral of five π‘ to the fourth power minus four π‘ cubed π’ minus four π‘ cubed plus eight π£ plus five π‘ cubed plus two π€ with respect to π‘.

Weβre asked to evaluate the integral. And we can see that this is the integral of a vector-valued function. We know this is a vector-valued function because it contains the unit directional vectors π’, π£, and π€. And itβs worth pointing out we know these are vectors because they use the hat notation, which represents unit vectors. However, sometimes weβll see vectors written with underlines, in bold notation, or with half arrows or full arrows. It doesnβt matter which notation is used; they all mean the same thing.

So we need to evaluate the integral of a vector-valued function. And to do this, we need to evaluate this component-wise. In other words, weβre going to integrate each of our component functions. Letβs start by integrating our first component function. Thatβs the integral of five π‘ to the fourth power minus four π‘ cubed with respect to π‘. And since this is the integral of a polynomial, we can do this term by term by using the power rule for integration.

We recall the power rule for integration tells us for any real constants π and π, where π is not equal to negative one, the integral of π times π‘ to the πth power with respect to π‘ is equal to π times π‘ to the power of π plus one divided by π plus one plus a constant of integration πΆ. We add one to our exponent and then divide by this new exponent.

So we can use this to evaluate our integral. In our first term, our exponent is four. So we add one to this to get five and then divide by five. This gives us five π‘ to the fifth power divided by five. And of course, we can simplify this. Five divided by five is equal to one. So the integral of our first term simplifies to give us π‘ to the fifth power. We can do the same for our second term. We add one to our exponent of three to give us four and then divide by this new exponent. This gives us negative four π‘ to the fourth power divided by four. And negative four divided by four simplifies to give us negative one. So the integral of our second term simplified to give us negative π‘ to the fourth power.

So integrating this, we got π‘ to the fifth power minus π‘ to the fourth power. And remember, we need to add a constant of integration. Weβll call this π. We now want to do the same to evaluate the integral of our second component function. However, we notice in our vector-valued function, weβre subtracting this vector. This, in fact, means we have two options. We could distribute negative one over our parentheses and then evaluate the integral as we usually would. However, we could also just evaluate the integral of four π‘ cubed plus π and then subtract this integral at the end. Either method works, and itβs personal preference which you use.

In this video, weβll just evaluate the integral four π‘ cubed plus eight with respect to π‘. And once again, this is the integral of a polynomial, so we can do this term by term by using the power for integration. In our first term, we get four π‘ to the fourth power divided by four. And we cancel the shared factor of four in our numerator and our denominator. This leaves us with π‘ to the fourth power. We could evaluate our second term in the same way. Instead of writing eight, we would write eight π‘ to the zeroth power. Remember, π‘ to the zeroth power is equal to one.

We could then apply the power rule of integration to this term to get eight π‘ to the first power divided by one. However, we can also just notice the derivative of eight π‘ with respect to π‘ is eight, so eight π‘ is an antiderivative of eight.

Finally, remember, we need to add a constant of integration. Weβll call this π. This gives us π‘ to the fourth power plus eight π‘ plus π. Now, we need to evaluate the integral of our third component function. Thatβs the integral of five π‘ cubed plus two with respect to π‘. And just as we did before, we can evaluate this term by term by using the power rule for integration. We get five π‘ to the fourth power over four plus two π‘ plus our constant of integration lowercase π. Weβre now ready to evaluate the integral of the vector-valued function given to us in the question.

We found the integral of our first component function to be π‘ to the fifth power minus π‘ to the fourth power plus π. So when we integrate our vector-valued function, weβll get π‘ to the fifth power minus π‘ to the fourth power plus π times π’. Similarly, when we integrated our second component function, we got π‘ to the fourth power plus eight π‘ plus π. But remember, this time, we need to subtract this vector. So weβre subtracting π‘ to the fourth power plus eight π‘ plus π times π£. And finally, when we integrated our third and final component function, we got five π‘ to the fourth power over four plus two π‘ plus lowercase π. So we need to add five π‘ to the fourth power over four plus two π‘ plus lowercase π times π€.

And we could leave our answer like this. However, thereβs one more piece of simplification we could do. We need to consider what would happen if we were to expand all of our parentheses. If we did this, we can see we would get π times π’ plus π times π£ plus π times π€. But remember, π, π, and π are constants. So ππ’ plus ππ£ plus ππ€ is a constant vector, so we can combine this into one constant vector, which we will call capital πΆ. And doing this gives us the following expression.

And this is our final answer. Therefore, we were able to evaluate the integral of the vector-valued function given to us in the question by doing this component-wise. We got π‘ to the fifth power minus π‘ to the fourth power π’ minus π‘ to the fourth power plus eight π‘ π£ plus five over four π‘ to the fourth power plus two π‘ π€ plus πΆ.