Video: Finding the Integration of a Function Using the Power Rule with Roots

Determine ∫ (9√(π‘₯) + 2π‘₯ βˆ’ (1/√π‘₯)) dπ‘₯.

04:50

Video Transcript

Determine the integral of nine times the square root of π‘₯ plus two π‘₯ minus one over the square root of π‘₯ with respect to π‘₯.

In this question, we’re asked to evaluate the integral of the sum or difference of three terms, and we know we can do this term by term. We know how to integrate two π‘₯ by using the power rule for integration. However, our other two terms are not too simple. We can’t evaluate the integral of these two terms directly. This means we’re going to need to rewrite these two terms in a form which we can integrate. We’ll do this by using our laws of exponents.

The first thing we need to notice is we can rewrite the square root of π‘₯ as π‘₯ to the power of one-half by using our laws of exponents. This will immediately allow us to integrate our first term. In fact, we’ll also use this to rewrite our third term. So using this, we’ve rewritten our integral as the integral of nine π‘₯ to the power of one-half plus two π‘₯ minus one divided by π‘₯ to the power of one-half with respect to π‘₯. And now, we can see we can evaluate the integral of our first term by using the power rule of integration and our second term. So all we need to do now is rewrite our third term in a form which we can integrate.

And in fact, the third term 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 rewrite our π‘₯-term in the numerator. And to do this, we just need to recall that one over π‘₯ to the 𝑛th power is equal to π‘₯ to the power of negative 𝑛 for any real constant 𝑛. In our case, the value of 𝑛 is one-half. So, we can rewrite the third term in our integrand as negative one times π‘₯ to the power of negative one-half.

This means we’ve now rewritten our integral as the integral of nine π‘₯ to the power of one-half plus two π‘₯ minus π‘₯ to the power of negative one-half with respect to π‘₯. And now, we’ve written this in a form we can integrate term by term 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. In other words, we add one to our exponent of π‘₯ and then divide by this new exponent.

We want to apply this term by term. Let’s start with the first term of our integrand. We can see our exponent of π‘₯ is equal to one-half. So, we want to add one to our exponent of one-half. This gives us a new exponent of three over two. And then, we need to divide by our new exponent of three over two. This gives us nine π‘₯ to the power three over two all divided by three over two. And there’s a few different ways of simplifying this. We’ll do this by multiplying both of our numerator and our denominator through by two.

In our denominator, we get two divided by two is equal to one. And in our numerator, two multiplied by nine is equal to 18. So, we simplified this to give us 18π‘₯ to the power of three over two divided by three. And of course, 18 divided by three is equal to six. So, by integrating our first term by using the power rule for integration and simplifying, we got six π‘₯ to the power of three over two.

We now want to do the same for our second term. To use the power rule for integration, we need to know the exponent of π‘₯. And to do this, we need to recall π‘₯ is the same as π‘₯ to the first power. So once again, to integrate this, we need to add one to our exponent then divide by this new exponent. Adding one to our exponent gives us a new exponent of two. And we then need to divide through by two. And of course, we can simplify this. Two divided by two is equal to one. So by integrating our second term by using the power rule for integration, this simplified to give us π‘₯ squared.

Finally, we need to apply the same process to our third and final term. We need to add one to our exponent of π‘₯ to the power of negative one-half. Adding one to negative one-half gives us a new exponent of one-half. Then, we need to divide by this new power of one-half. So, we’re subtracting π‘₯ to the power of one-half divided by one-half. And we know dividing by one-half is the same as multiplying by two. So, integrating our third term by using the power rule for integration gave us negative two times π‘₯ to the power of one-half. And remember, we need to add our constant of integration 𝐢. And we could leave our answer like this. However, we’ll use our laws of exponents to simplify.

First, recall, we can rewrite π‘₯ to the power of three over two as π‘₯ cubed all raised to the power of one-half. But remember, raising a number to the power of one-half is the same as taking the square root. So, in fact, π‘₯ raised to the power three over two is the same as the square root of π‘₯ cubed. So, we’ll rewrite our first term as six times the square root of π‘₯ cubed. We’ll also rewrite our third term as negative two times the square root of π‘₯. This gives us six times the square root of π‘₯ cubed plus π‘₯ squared minus two root π‘₯ plus 𝐢. And the last piece of simplification we’ll do is rearrange our third and first term, and doing this gives us our final answer.

Therefore, we were able to show the integral of nine root π‘₯ plus two π‘₯ minus one over the square root of π‘₯ with respect to π‘₯ is equal to negative two root π‘₯ plus π‘₯ squared plus six times the square root of π‘₯ cubed plus our constant of integration 𝐢.

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