Question Video: Finding the Antiderivative of a Polynomial Function Mathematics • Higher Education

Determine the antiderivative 𝐹 of the function 𝑓(π‘₯) = 5π‘₯⁴ + 4π‘₯Β³ where 𝐹(1) = βˆ’2.

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Video Transcript

Determine the antiderivative capital 𝐹 of the function lowercase 𝑓 of π‘₯ equals five π‘₯ to the fourth power plus four π‘₯ cubed, where capital 𝐹 of one equals negative two.

The general antiderivative of a function lowercase 𝑓 of π‘₯ is the function capital 𝐹 of π‘₯ plus 𝐢 such that the first derivative of capital 𝐹 of π‘₯, 𝐹 prime of π‘₯, is equal to 𝑓 of π‘₯ and 𝐢 is any real constant. An antiderivative is not unique, and there are many functions which differ up to a constant which have the same derivative. In this instance though, we’ve been given some more information. The value of the antiderivative 𝐹 when π‘₯ is equal to one is negative two. And so we will be able to determine the value of 𝐢 that gives a unique antiderivative satisfying this boundary condition. The function 𝑓 of π‘₯ in this question is a polynomial. It is the sum of two terms which are each constants multiplied by powers of π‘₯.

An antiderivative is linear, so the antiderivative of a sum is the sum of the antiderivatives. And we can therefore find the antiderivatives of each term separately and add them together. Antidifferentiation is the reverse process of differentiation. Recalling the power rule of differentiation, we know that the first derivative of π‘₯ to the power of π‘Ž plus one over π‘Ž plus one is equal to π‘₯ to the power of π‘Ž, provided π‘Ž is not equal to negative one. So working in reverse, the general antiderivative of π‘₯ to the π‘Žth power is π‘₯ to the power of π‘Ž plus one over π‘Ž plus one plus the constant of antidifferentiation 𝐢. And again π‘Ž must be not equal to negative one for this result to hold.

It also follows that the antiderivative of a constant multiplied by 𝑓 of π‘₯ is just that constant multiplied by the antiderivative. Applying these results to the first term in 𝑓 of π‘₯, we find that the antiderivative of five π‘₯ to the fourth power is five π‘₯ to the fifth power over five plus a constant 𝐢 one. And applying the same results to the second term gives the antiderivative four π‘₯ to the fourth power over four plus a constant 𝐢 two. We can then simplify this to π‘₯ to the fifth power plus π‘₯ to the fourth power. And the two constants of antidifferentiation 𝐢 one and 𝐢 two can be combined into a single constant 𝐢.

Now, this is the most general antiderivative 𝐹 of the function 𝑓 of π‘₯. But remember, we were given a boundary condition. We can use this condition to determine the constant 𝐢, which gives a unique antiderivative satisfying 𝐹 of one equals negative two. Substituting π‘₯ equals one and 𝐹 of π‘₯ equals negative two, we have the equation negative two equals one to the fifth power plus one to the fourth power plus 𝐢. One to the fifth power and one to the fourth power are each one. So subtracting two from each side, we find that the value of 𝐢 is negative four.

Substituting this value of 𝐢 back into our function 𝐹 of π‘₯, and we have our answer. The unique antiderivative 𝐹 of the function 𝑓 of π‘₯ equals five π‘₯ to the fourth power plus four π‘₯ cubed satisfying 𝐹 of one equals negative two is 𝐹 of π‘₯ equals π‘₯ to the fifth power plus π‘₯ to the fourth power minus four.

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