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

Calculate the integral ∫ [(4𝑑³ + 3𝑑²) 𝐒 +(4𝑑² βˆ’ 5) 𝐣 + (4𝑑³ βˆ’ 5𝑑² + 3) 𝐀] d𝑑.

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Video Transcript

Calculate the indefinite integral of four 𝑑 cubed plus three 𝑑 squared 𝐒 plus four 𝑑 squared minus five 𝐣 plus four 𝑑 cubed minus five 𝑑 squared plus three 𝐀 with respect to 𝑑.

This is a vector-valued function. This takes a real number 𝑑. And it outputs a position vector. And to integrate a vector-valued function, we simply integrate each component in the usual way. So we’ll integrate the component for 𝐒 with respect to 𝑑. That’s four 𝑑 cubed plus three 𝑑 squared. We’ll integrate the component for 𝐣. That’s four 𝑑 squared minus five. And we’ll integrate the component for 𝐀, four 𝑑 cubed minus five 𝑑 squared plus three with respect to 𝑑. 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 𝑑 cubed is four 𝑑 to the fourth power over four. Integrating three 𝑑 squared and we obtain three 𝑑 cubed over three. And of course, this is an indefinite integral. So we must have that constant of integration π‘Ž. Simplifying fully and we obtain 𝑑 to the fourth power plus 𝑑 cubed plus π‘Ž.

Similarly, when we integrate four 𝑑 squared, we get four 𝑑 cubed over three. The integral of negative five is negative five 𝑑. And then, we need another constant of integration 𝑏. In our final component, when we integrate four 𝑑 cubed, once again, we get four 𝑑 to the fourth power over four. The integral of negative five 𝑑 squared is negative five 𝑑 cubed over three. And the integral of three is three 𝑑. Let’s have the final constant of integration 𝑐. This simplifies as shown.

We put this back into vector form. And we see that our integral is equal to 𝑑 to the fourth power plus 𝑑 cubed plus π‘Ž 𝐒 plus four-thirds 𝑑 cubed minus five 𝑑 plus 𝑏 𝐣 plus 𝑑 to the fourth power minus five-thirds 𝑑 cubed plus three 𝑑 plus 𝑐 𝐀. Notice, though, that each of our components has its own constant. So we can combine these and form a constant vector 𝐜. And this means our integral is equal to 𝑑 to the fourth power plus 𝑑 cubed 𝐒 plus four-thirds 𝑑 cubed minus five 𝑑 𝐣 plus 𝑑 to the fourth power minus five-thirds 𝑑 cubed plus three 𝑑 𝐀 plus this capital 𝐂, which represents a constant vector.

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