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

Determine ∫ (the fifth root of 𝑦²)/(the cube root of 𝑦) d𝑦.

03:47

Video Transcript

Determine the indefinite integral of the fifth root of 𝑦 squared divided by the cube root of 𝑦 with respect to 𝑦.

In this question, we’re asked to evaluate the integral of the quotient of two functions. And our integrand is given in terms of 𝑦, and we’re integrating with respect to 𝑦. Remember, as long as our integrand and what we’re integrating with respect to share the same variable, it doesn’t matter what we call this variable. So, we can integrate this using all of our standard rules. However, as it stands, we don’t know how to integrate this function directly. So, we’re going to need to rewrite this in a form which we can integrate. And since this expression involves a lot of exponents, we’ll do this by using our laws of exponents.

The first one of our laws of exponents we’ll use is the following. We know the 𝑛th root of π‘₯ is equal to π‘₯ to the power of one over 𝑛. Of course, this is only true for positive integers 𝑛. And in fact, we can use this to rewrite both the numerator of our integrand and the denominator. First, in our numerator, the value of 𝑛 is five. So, we get 𝑦 squared all raised to the power of one over five. Then, in our denominator, we can rewrite the cube root of 𝑦 as 𝑦 to the power of one over three.

So, we’ve rewritten our integral in the following form. However, we still can’t evaluate this integral. So, we’re going to need to simplify this even further. We can see in our numerator we have 𝑦 squared all raised to the power of one over five. And we can rewrite this as 𝑦 raised to the power of some number by using our laws of exponents. Recall, π‘Ž raised to the power of 𝑏 all raised to the power of 𝑐 is equal to π‘Ž raised to the power of 𝑏 times 𝑐. We’ll apply this to rewrite the numerator of our integrand. Our value of π‘Ž is 𝑦, our value of 𝑏 is two, and our value of 𝑐 is one over five. This gives us a new numerator of 𝑦 to the power of two times one over five.

So, now, we have the integral of 𝑦 to the power of two times one over five divided by one to the power of one-third with respect to 𝑦. And of course, we can simplify the exponents in our numerator. Two times one over five is equal to two over five. Now, we can see our integrand is the quotient of two exponents where 𝑦 is the base in both of these. And once again, we can simplify this by using our laws of exponents. We know 𝑦 raised to the power of π‘Ž divided by 𝑦 raised to the power of 𝑏 is equal to 𝑦 raised to the power of π‘Ž minus 𝑏.

In this case, our value of π‘Ž is two over five and our value of 𝑏 is one-third. So using this, we get the integral of 𝑦 over two over five minus one-third with respect to 𝑦. And we can simplify this. Evaluating our exponent, we get two over five minus one-third is equal to one over 15. So, after all of this manipulation, we were able to rewrite our integral as the integral of 𝑦 raised to the power of one over 15 with respect to 𝑦. And now, we can evaluate this integral by using the power rule for integration. Recall for all real constants π‘Ž and 𝑛, where 𝑛 is not equal to negative one, the integral of π‘Ž times 𝑦 to the 𝑛th power with respect to 𝑦 is equal to π‘Ž times 𝑦 to the power of 𝑛 plus one divided by 𝑛 plus one plus our constant of integration 𝐢.

So, we want to add one to our exponent of 𝑦 and then divide by this new exponent. We can see in our case the exponent of 𝑦 is one over 15. So, we need to add one to this exponent of one over 15 and then divide by this new exponent. This gives us 𝑦 to the power of one over 15 plus one all divided by one over 15 plus one. And of course, we need to add our constant of integration 𝐢.

And the last thing we’ll do is simplify our exponent and our denominator. One over 15 plus one is equal to 16 over 15. This gives us 𝑦 to the power of 16 over 15 divided by 16 over 15 plus 𝐢. And we can do one last piece of simplification. Instead of dividing by 16 over 15, we’ll multiply by the reciprocal. And doing this gives us our final answer.

Therefore, we were able to show the integral of the fifth root of 𝑦 squared over the cube root of 𝑦 with respect to 𝑦 is equal to 15 times 𝑦 to the power of 16 over 15 divided by 16 plus 𝐢.

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