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

Determine ∫ 8/(the fifth root of π‘₯⁢) dπ‘₯.

03:19

Video Transcript

Determine the integral of eight divided by the fifth root of π‘₯ to the sixth power with respect to π‘₯.

In this question, we’re asked to evaluate the indefinite integral of a radical. And we can see we can’t evaluate this integral directly in its following form. However, we can use our laws of exponents to rewrite this in a form which we can integrate.

First, we want to simplify the expression in our denominator. That’s the fifth root of π‘₯ to the sixth power. We can do this by recalling the 𝑛th root of π‘Ž is equal to π‘Ž to the power of one over 𝑛. In our case, the value of 𝑛 is five and the value of π‘Ž is π‘₯ to the sixth power. So we can rewrite the denominator of our integrand as π‘₯ to the sixth power raised to the power of one over five.

However, we still can’t evaluate this integral directly. We can see in our denominator we have π‘₯ to the sixth power all raised to the power of one-fifth. We can rewrite this by using one of our laws of exponents. This time, we need to recall that π‘Ž raised to the power of 𝑏 all raised to the power of 𝑐 is the same as π‘Ž raised to the power of 𝑏 times 𝑐. We can apply this by setting our value of π‘Ž equal to π‘₯, 𝑏 equal to six, and 𝑐 equal to one-fifth. This lets us rewrite the denominator of our integrand as π‘₯ to the power of six times one-fifth. And, of course, we can simplify our exponent. Six times one-fifth is equal to six over five.

So now we’ve rewritten our integral as the integral of eight divided by π‘₯ to the power of six over five with respect to π‘₯. And this is almost in a form we can integrate by using the power rule for integration. We just need to use our laws of exponents to bring the π‘₯-term into our numerator. And we can do this by using one more of our laws of exponents.

We recall dividing by π‘₯ to the power of 𝑏 is the same as multiplying by π‘₯ to the power of negative 𝑏. So by using π‘Ž is equal to eight and 𝑏 is equal to six over five, we can rewrite our integral as the integral of eight times π‘₯ to the power of negative six over five with respect to π‘₯. And now we’ve rewritten our integral in a form we can evaluate by using the power rule for integration.

We recall this tells us for any real constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Ž times π‘₯ to the 𝑛th power with respect to π‘₯ is equal to π‘Ž times π‘₯ to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐢. We add one to our exponent of π‘₯ and then divide by this new exponent. In our case, the exponent of π‘₯ is negative six over five. So first, we need to add one to our exponent of negative six over five. This gives us negative six over five plus one. And we can evaluate this. It’s equal to negative one over five. So our new exponent is negative one-fifth, and this tells us we need to divide by negative one-fifth.

Finally, we need to add our constant of integration 𝐢. And we can simplify this. Instead of dividing by negative one-fifth, we can multiply by the reciprocal of negative one-fifth. This gives us negative five multiplied by eight times π‘₯ to the power of negative one-fifth plus 𝐢. And we can just evaluate this. Negative five multiplied by negative eight is equal to negative 40. This gives us negative 40π‘₯ to the power of negative one-fifth plus 𝐢.

And it’s worth pointing out we could use our laws of exponents to bring π‘₯ into the denominator and write this as the fifth root of π‘₯. It won’t change the value of our answer. However, we can do this. It’s all personal preference. We’ll leave our answer as it is.

Therefore, we were able to show the integral of eight divided by the fifth root of π‘₯ to the sixth power with respect to π‘₯ is equal to negative 40 times π‘₯ to the power of negative one over five plus 𝐢.

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