# Video: Finding the Set on Which a Piecewise-Defined Function Involving Trigonometric Ratios Is Continuous

Calculate the integral β« [(4π‘Β³ + 3π‘Β²) π’ +(4π‘Β² β 5) π£ + (4π‘Β³ β 5π‘Β² + 3) π€] dπ‘.

02:18

### Video Transcript

Calculate the indefinite integral of four π‘ cubed plus three π‘ squared π’ plus four π‘ squared minus five π£ plus four π‘ cubed minus five π‘ squared plus three π€ with respect to π‘.

This is a vector-valued function. This takes a real number π‘. And it outputs a position vector. And to integrate a vector-valued function, we simply integrate each component in the usual way. So weβll integrate the component for π’ with respect to π‘. Thatβs four π‘ cubed plus three π‘ squared. Weβll integrate the component for π£. Thatβs four π‘ squared minus five. And weβll integrate the component for π€, four π‘ cubed minus five π‘ squared plus three with respect to π‘. These are polynomial functions. And we know that, to integrate a polynomial term whose exponent is not equal to negative one, we increase that exponent by one and then divide by that number. This means the integral of four π‘ cubed is four π‘ to the fourth power over four. Integrating three π‘ squared and we obtain three π‘ cubed over three. And of course, this is an indefinite integral. So we must have that constant of integration π. Simplifying fully and we obtain π‘ to the fourth power plus π‘ cubed plus π.

Similarly, when we integrate four π‘ squared, we get four π‘ cubed over three. The integral of negative five is negative five π‘. And then, we need another constant of integration π. In our final component, when we integrate four π‘ cubed, once again, we get four π‘ to the fourth power over four. The integral of negative five π‘ squared is negative five π‘ cubed over three. And the integral of three is three π‘. Letβs have the final constant of integration π. This simplifies as shown.

We put this back into vector form. And we see that our integral is equal to π‘ to the fourth power plus π‘ cubed plus π π’ plus four-thirds π‘ cubed minus five π‘ plus π π£ plus π‘ to the fourth power minus five-thirds π‘ cubed plus three π‘ plus π π€. Notice, though, that each of our components has its own constant. So we can combine these and form a constant vector π. And this means our integral is equal to π‘ to the fourth power plus π‘ cubed π’ plus four-thirds π‘ cubed minus five π‘ π£ plus π‘ to the fourth power minus five-thirds π‘ cubed plus three π‘ π€ plus this capital π, which represents a constant vector.