# Question Video: Calculating an Integral of a Vector-Valued Function Mathematics • Higher Education

Evaluate the integral β«[3π‘Β² π’ + 5π‘β΄ π£ + 4π‘β΅ π€] dπ‘.

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### 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‘.

Weβre 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 ππ‘ to the πth 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 weβve 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 ππ’ plus ππ£ plus ππ€ 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.