Question Video: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents | Nagwa Question Video: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents | Nagwa

Question Video: Finding the General Antiderivative of a Function Using the Power Rule of Integration with Fractional Exponents Mathematics

Determine the most general antiderivative 𝐹(π‘₯) of the function 𝑓(π‘₯) = βˆ’2 the cubed root of (π‘₯Β²) + 3π‘₯√π‘₯.

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Video Transcript

Determine the most general antiderivative capital 𝐹 of π‘₯ of the function lowercase 𝑓 of π‘₯ equal to negative two times the cubed root of π‘₯ squared plus three π‘₯ times root π‘₯.

We begin by noting the different notations used in the question. We have a function lowercase 𝑓 of π‘₯ and then a function capital 𝐹 of π‘₯, which represents its antiderivative. The general antiderivative of a function lowercase 𝑓 of π‘₯ is the function capital 𝐹 of π‘₯ plus 𝑐 such that the first derivative of 𝐹 of π‘₯, 𝐹 prime of π‘₯, is equal to 𝑓 of π‘₯ and 𝑐 is any real constant. In other words, an antiderivative of a function lowercase 𝑓 is another function capital 𝐹, whose derivative is equal to the original function.

An antiderivative is not unique, and there are many functions which differ up to a constant which have the same derivative. The function 𝑓 of π‘₯ here 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. We can therefore find the antiderivatives of each term separately. The process of finding an antiderivative is the reverse process of differentiating a function.

Let’s simplify the function itself first. And to do this, we’ll need to recall some laws of exponents. To simplify the first term, we should recall the 𝑛th roots to find reciprocal exponents. The 𝑛th root of π‘₯ is π‘₯ to the power of one over 𝑛. And so the cubed root of π‘₯ squared is π‘₯ squared to the power of one-third. We should also recall the power rule of exponents. If we raise a base to a power and then another power, we multiply the powers together. So π‘₯ squared to the power of one-third is π‘₯ to the power of two-thirds.

For the second term, we recall another law of exponents. If we multiply together two powers of the same base, then we add the exponents. So π‘₯ multiplied by root π‘₯, which we can write as π‘₯ to the power of one-half, is π‘₯ to the power of one plus a half, or π‘₯ to the power of three over two. The function 𝑓 of π‘₯ can therefore be written as negative two π‘₯ to the power of two-thirds plus three π‘₯ to the power of three over two.

Next, we recall 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 the general antiderivative of π‘₯ to the power of π‘Ž, or π‘₯ to the π‘Žth power, is π‘₯ to the power of π‘Ž plus one over π‘Ž plus one plus a constant of antidifferentiation 𝑐, provided again that π‘Ž is not equal to negative one. We also know that the antiderivative of π‘Ž multiplied by 𝐹 of π‘₯ is π‘Ž multiplied by 𝐹 prime of π‘₯ or π‘Ž multiplied by the function lowercase 𝑓 of π‘₯, for real constants π‘Ž. In other words, if a derivative is multiplied by a constant, the antiderivative is multiplied by that same constant and vice versa. So for our function 𝑓 of π‘₯, we can find the antiderivative of each power of π‘₯ and then multiply by the constants of negative two and positive three.

Let’s do this term by term then. Applying these results to the first term, we have that the antiderivative of negative two π‘₯ to the power of two-thirds is negative two π‘₯ to the power of two-thirds plus one over two-thirds plus one plus a constant of antidifferentiation 𝑐 one. Of course, two-thirds plus one is two-thirds plus three-thirds, or five-thirds. And then we can recall that dividing by five-thirds is equivalent to multiplying by three-fifths. So the antiderivative of the first term is negative six over five π‘₯ to the power of five-thirds plus 𝑐 one.

Applying these results to the second term, we have that the antiderivative of three π‘₯ to the power of three over two is three π‘₯ to the power of three over two plus one over three over two plus one plus another constant of antidifferentiation 𝑐 two. Three over two plus one is five over two, and dividing by five over two is equivalent to multiplying by two-fifths. So this term simplifies to six π‘₯ to the power of five over two over five plus a constant of antidifferentiation 𝑐 two.

Finally, we can combine these two constants of antidifferentiation into a single constant which we’ll call 𝑐. And so we have our answer to the problem. The most general antiderivative capital 𝐹 of π‘₯ of the function lowercase 𝑓 of π‘₯ is negative six π‘₯ to the power of five over three over five plus six π‘₯ to the power of five over two over five plus a constant of antidifferentiation 𝑐.

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