### Video Transcript

Evaluate the integral the integral of three π‘ squared π’ plus five π‘ to the fourth power π£ plus four π‘ to the fifth power π€ with respect to π‘.

Weβre asked to calculate the integral of a vector-valued function. When we input a value of π‘, our output is a position vector. To evaluate the integral of a vector-valued function, we integrate each component separately. We start by integrating the first component, three π‘ squared with respect to π‘. And we can integrate this by using the power rule for integration. Which tells us for constants π and π, where π is not equal to negative one, to integrate ππ‘ to the πth power with respect to π‘, we add one to the exponent and then divide by the new exponent. Then, we add our constant of integration π. We add one to our exponent and then divide by the new exponent.

This gives us three π‘ cubed over three plus a constant of integration we will call π. We can then cancel the shared factor of three to get π‘ cubed plus π. Now, we need to integrate the second term, five π‘ to the fourth power with respect to π‘. We also integrate this by using the power rule for integration. This gives us five π‘ to the fifth power over five plus a constant of integration we will call π. And we can cancel the shared factor of five to get π‘ to the fifth power plus π.

We now want to integrate the last part of our vector-valued function, four π‘ to the fifth power with respect to π‘. Again, we integrate this by adding one to the exponent and dividing by the new exponent. This gives us four π‘ to the sixth power divided by six plus a constant of integration we will call π. And we can simplify this to get two π‘ to the sixth power over three plus π. Since weβve now found the integral of each of our component functions, we can conclude that the integral of our vector-valued function is given by π‘ cubed plus π π’ plus π‘ to the fifth power plus π π£ plus two-thirds π‘ to the sixth power plus π π€.

We could leave our answer like this. However, we see the π, π, and π are constants. So if we were to expand and rearrange our answer, we would see that ππ’ plus ππ£ plus ππ€ is just a position vector. We could call this π, giving us that the integral of our vector-valued function with respect to π‘ is equal to π‘ cubed π’ plus π‘ to the fifth power π£ plus two over three π‘ to the sixth power π€ plus π.