Lesson Video: Indefinite Integrals: The Power Rule Mathematics • 12th Grade

Video Transcript

Indefinite Integrals: The Power Rule

In this video, we will learn how to find the indefinite integrals of polynomials and general power functions using the power rule for integration. Let’s start by recalling what the antiderivative of a function is. We can say that capital 𝐹 is the antiderivative of lowercase 𝑓 if capital 𝐹 prime of 𝑥 is equal to lowercase 𝑓 of 𝑥. We can in fact say that this is true for any function 𝑔 of 𝑥 where 𝑔 of 𝑥 is equal to capital 𝐹 of 𝑥 plus 𝐶 for any constant 𝐶. Now, this is very useful as we’ll be using it to define an indefinite integral.

We can say that the indefinite integral of lowercase 𝑓 of 𝑥 with respect to 𝑥 is equal to capital 𝐹 of 𝑥 plus 𝐶 where capital 𝐹 is the antiderivative of lowercase 𝑓. It’s very important to remember our constant of integration when performing an indefinite integral. Let’s quickly consider why this constant is here. If we perform the reverse operation on this equation, so that’s differentiating with respect to 𝑥, then since we are simply performing the inverse operation on the left, the differential with respect to 𝑥 of the integral of lowercase 𝑓 of 𝑥 with respect to 𝑥 will just be lowercase 𝑓 of 𝑥.

On the right-hand side, when we differentiate capital 𝐹 of 𝑥 with respect to 𝑥, we get 𝐹 prime of 𝑥 and our constant 𝐶 will simply vanish because differentiating a constant gives zero. Now, when we go back the other way to our integral, the constant will appear again. However, we do not know the value of this constant, hence why we label it 𝐶. It’s just an unknown constant. Let’s now consider what happens when we integrate some function with respect to 𝑥, for example, three 𝑥.

We know the equation for finding an indefinite integral. We have that the integral of lowercase 𝑓 of 𝑥 with respect to 𝑥 is equal to uppercase 𝐹 of 𝑥 plus 𝐶 where uppercase 𝐹 is the antiderivative of lowercase 𝑓. Now in our case, lowercase 𝑓 of 𝑥 is equal to three 𝑥. So we simply need to find the antiderivative of three 𝑥. Let’s try and do this by trial and error. We’re trying to work out what we need to differentiate in order to get three 𝑥. Let’s start by differentiating 𝑥 squared with respect to 𝑥. This gives us two 𝑥 which is very close to three 𝑥. However, we’re not quite there yet.

Let’s try multiplying our differential by one-half. Differentiating 𝑥 squared over two with respect to 𝑥 give us 𝑥 which is one-third of what we’re trying to find. So let’s now try multiplying our differential by three. And differentiating three 𝑥 squared over two with respect to 𝑥 gives us three 𝑥. Therefore, we found our antiderivative. And it’s three 𝑥 squared over two. So this is all capital 𝐹 of 𝑥. Therefore, we can say that our integral is equal to three 𝑥 squared over two plus 𝐶. Now this is quite a lengthy method for finding antiderivatives of power functions.

Let’s consider the power rule for derivatives. We know that the derivative of 𝑥 to the power of 𝑛 plus one with respect to 𝑥 is 𝑛 plus one multiplied by 𝑥 to the power of 𝑛 since for the power rule of derivatives we multiply by the power and then decrease the power by one. What this tells us is that 𝑥 to the power of 𝑛 plus one is the antiderivative of 𝑛 plus one multiplied by 𝑥 to the power of 𝑛. We can therefore recall 𝑥 to the power of 𝑛 plus one capital 𝐹 of 𝑥 and 𝑛 plus one multiplied by 𝑥 to the power of 𝑛 lowercase 𝑓 of 𝑥. And we can substitute these functions into our equation for the indefinite integral of a function. And this gives us that the integral of 𝑛 plus one multiplied by 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one plus 𝐶.

Now, we have a constant inside our integral. And we can, in fact, factor this out. Let’s quickly consider why we can do this. We know that if we differentiate a constant multiplied by some function 𝑔 of 𝑥, then this is equal to the constant multiplied by the differential of 𝑔 of 𝑥 with respect to 𝑥. This, therefore, tells us that the integral of 𝑎 multiplied by 𝑔 of 𝑥 with respect to 𝑥 is equal to 𝑎 multiplied by the integral of 𝑔 of 𝑥 with respect to 𝑥. Therefore we’re able to factor out our constant of 𝑛 plus one. And then we can simply divide by this constant. And this gives us that the integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶 over 𝑛 plus one.

However, since both 𝐶 and 𝑛 plus one are constants, this means that 𝐶 over 𝑛 plus one will also be a constant. And we can call that constant 𝐷. Therefore, we arrive at our power rule for integration. It tells us that the integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶. Here I have relabeled our constant of integration back to 𝐶 just for consistency with our definition of an indefinite integral. However, it does not matter what you label this constant. An easy way to remember this rule is that when we integrate a power function, we increase the power by one and then divide by the new power. And of course, we mustn’t forget to add our constant of integration.

Now, this power rule for integration works for any real value of 𝑛 except for one specific value. And that is when 𝑛 is equal to negative one. If we tried using 𝑛 is equal to negative one, we get that the indefinite integral of 𝑥 to the power of negative one with respect to 𝑥 is equal to 𝑥 to the power of negative one plus one over negative one plus one plus 𝐶. Since negative one plus one is zero, this means that we have a zero in the denominator of our fraction. Therefore, this fraction is undefined and so is the integral. And so we can say that our power rule for integration works for all real values of 𝑛 except negative one. Now it is, in fact, possible to integrate 𝑥 to the power of negative one with respect to 𝑥. However, we will not be covering it in this video. We will now be moving on to an example of how we can use the power rule.

Determine the indefinite integral of negative 𝑥 to the power of nine with respect to 𝑥.

In order to find this integral, we need to use the power rule for integration. We have that the integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶. And we’re trying to find the integral of negative 𝑥 to the power of nine with respect to 𝑥. Let’s start by factoring out the negative one. We have that our integral is equal to the negative integral of 𝑥 to the power of nine with respect to 𝑥. Now, when we look at our integral and we compare it to the integral in the formula, we can see that in our case 𝑛 is equal to nine. Therefore, we simply substitute 𝑛 is equal to nine into the formula. Therefore, we obtain negative 𝑥 to the power of nine plus one all over nine plus one plus 𝐶. This gives negative 𝑥 to the power of 10 over 10 minus 𝐶.

Now, negative 𝐶 is just another constant. So we can call it 𝐷. And now we’ve reached our solution. And that is that the indefinite integral of negative 𝑥 to the power of nine with respect to 𝑥 is equal to negative 𝑥 to the power of 10 over 10 plus 𝐷.

Let’s now work out how we will integrate a polynomial. We know that the differential of a sum of functions, so 𝑓 of 𝑥 plus 𝑔 of 𝑥 with respect to 𝑥, is equal to the sum of the differential of the functions. So 𝑑 by d𝑥 of 𝑓 of 𝑥 plus 𝑑 by d𝑥 of 𝑔 of 𝑥. From this we obtain that the integral of a sum of functions is equal to the sum of the integrals of the functions.

Using this rule, we’re able to split integrals of polynomials down into integrals of power functions. For example, the integral of 𝑥 squared plus 𝑥 with respect to 𝑥 is equal to the integral of 𝑥 squared with respect to 𝑥 plus the integral of 𝑥 with respect to 𝑥. And we already know how to integrate these power functions on the right. Using this, we’re able to integrate any polynomial. Let’s now look at an example.

Determine the indefinite integral of 25𝑥 squared minus 65𝑥 plus 36 with respect to 𝑥.

Let’s start by splitting this integral up using the fact that the integral of a sum of functions is equal to the sum of the integrals of the functions. We obtain that our integral is equal to the integral of 25𝑥 squared with respect to 𝑥 plus the integral of negative 65𝑥 with respect to 𝑥 plus the integral of 36 with respect to 𝑥. We can factor out the constant in each integral. Now, we can use the power rule for integration which tells us that the integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶.

In the case of our first integral, 𝑛 is two. Therefore, it’s equal to 25 multiplied by 𝑥 cubed over three plus some constant which we’ll call 𝐶 one. For our second integral, we’re integrating 𝑥. 𝑥 is equal to 𝑥 to the power of one. Therefore, 𝑛 is equal to one. Therefore, it’s equal to negative 65 multiplied by 𝑥 squared over two plus some constant which we’ll call 𝐶 two. For our third integral, we’re integrating one. One is also equal to 𝑥 to the power of zero. Therefore, our value of 𝑛 is zero. And so it’s equal to 36 multiplied by 𝑥 to the power of one over one plus some constant which we’ll call 𝐶 three.

Expanding and simplifying, we obtain that our integral is equal to 25𝑥 cubed over three minus 65𝑥 squared over two plus 36𝑥 plus 25𝐶 one minus 65𝐶 two plus 36𝐶 three. Now, since 𝐶 one, 𝐶 two, and 𝐶 three are all constant, 25𝐶 one minus 65𝐶 two plus 36𝐶 three is also a constant. And we can relabel this as 𝐶. And so we reach our solution. And that is that the indefinite integral of 25𝑥 squared minus 65𝑥 plus 36 with respect to 𝑥 is equal to 25𝑥 cubed over three minus 65𝑥 squared over two plus 36𝑥 plus 𝐶.

In the next example, we’ll see how we can simply integrate a polynomial without splitting it up into separate integrals.

Determine the indefinite integral of 𝑥 minus six multiplied by 𝑥 minus five multiplied by 𝑥 minus three.

Let’s start by expanding the brackets. Expanding the first two sets of brackets, we get 𝑥 squared minus 11𝑥 plus 30. And we then multiply this by 𝑥 minus three. We obtain the indefinite integral of 𝑥 cubed minus 14𝑥 squared plus 63𝑥 minus 90 which is in fact a polynomial. And we can use the power rule for integration to integrate this term by term. The power rule tells us that the integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶. If instead we were integrating 𝑥 to the power of 𝑛 multiplied by some constant 𝑎, then since we can factor a constant out of our integral, then this would simply be equal to 𝑎 multiplied by 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶.

Now, you may be wondering why we haven’t multiplied the 𝐶 by 𝑎. And that’s because 𝑎 is also a constant. Therefore, 𝑎 multiplied by 𝐶 is a constant. And we can just rename this new constant to be 𝐶. So now, let’s apply this rule to our integral term by term. The first term is 𝑥 cubed. Therefore, 𝑛 is equal to three. We increase the power by one and divide by the new power to get 𝑥 to the power of four over four. The next term is negative 14𝑥 squared. Negative 14 is just a constant. So that will remain. Our power is two. So 𝑛 is two. We increase the power by one to get 𝑥 cubed and divide by the new power.

On next term, we have 63𝑥. So we can start by writing in our constant of 63. We then note that 𝑥 is equal to 𝑥 to the power of one. So 𝑛 is equal to one. So when we integrate this, we get 𝑥 squared over two. For our final term, we have negative 90. And we know that this can also be written as negative 90 multiplied by 𝑥 to the power of zero since 𝑥 to the power of zero is just one. So now, we can integrate it. We start by writing our constant of negative 90. Since our power of 𝑥 is zero, we increase the power by one giving us 𝑥 to the power of one and divide by the new power. So that’s dividing by one. And we miss then adding our constant of integration 𝐶.

We can write this out a little neater for our solution which is 𝑥 to the power of four over four minus 14𝑥 cubed over three plus 63𝑥 squared over two minus 90𝑥 plus 𝐶.

Let’s quickly note that this integration power rule works for any real 𝑛 except negative one. So that includes both negative and noninteger powers of 𝑥.

Let’s see how this works in the following examples.

Determine the indefinite integral of negative two over seven multiplied by 𝑥 to the power of negative nine with respect to 𝑥.

In our integral, we simply have a power function. So we can use the power rule for integration in order to find this integral. It tells us that the indefinite integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶. In our case, we’re integrating negative two-sevenths 𝑥 to the power of negative nine with respect to 𝑥. So our power of 𝑥 is negative nine. We can start by writing down our constant which is negative two-sevenths.

Then since our value of 𝑛 is negative nine, we next need to write 𝑥 to the power of 𝑛 plus one over 𝑛 plus one. So that’s 𝑥 to the power of negative nine plus one which is 𝑥 to the power of negative eight over negative nine plus one. So that’s negative eight. Then we mustn’t forget to add our constant of integration 𝐶. For our final step here, we just need to simplify. And we obtain our solution which is that the indefinite integral of negative two-seventh multiplied by 𝑥 to the power of negative nine with respect to 𝑥 is equal to 𝑥 to the power of negative eight over 28 plus 𝐶.

In our final example, we’ll see how to integrate a function with noninteger powers of 𝑥.

Determine the indefinite integral having negative four multiplied by the fifth root of 𝑥 to the power of nine plus eight all multiplied by the fifth root of 𝑥 squared with respect to 𝑥.

Let’s start by writing this roots as powered. We know that the 𝑛th root of 𝑥 is equal to 𝑥 to the power of one over 𝑛. Once we’ve written our roots as powers, we can then combine them with the existing powers, using the fact that 𝑥 to the power of 𝑛 to the power of 𝑚 is equal to 𝑥 to the power of 𝑛 multiplied by 𝑚. Therefore, 𝑥 to the power of nine to the power of one-fifth becomes 𝑥 to the power of nine-fifths. And 𝑥 squared to the power of one-fifths becomes 𝑥 to the power of two-fifths. Now, we can expand the brackets, using the fact that 𝑥 to the power of 𝑛 times 𝑥 to the power of 𝑚 is equal to 𝑥 to the power of 𝑛 plus 𝑚. So our integral becomes the integral of negative four 𝑥 to the power of eleven-fifths plus eight multiplied by 𝑥 to the power of two-fifths with respect to 𝑥.

Here, we can use the power rule for integration which tells us that the indefinite integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶. We can apply this rule to our integral term by term. For the first term, we have negative four 𝑥 to the power of eleven-fifths. Therefore, 𝑛 is equal to eleven-fifths. When we integrate this term, we get negative four multiplied by 𝑥 to the power of 𝑛 plus one. And 𝑛 plus one is simply sixteen-fifths. So it’s 𝑥 to the power of sixteen-fifths. And then we need to divide by 𝑛 plus one. So that’s dividing by sixteen-fifths.

For the second time, we have eight multiplied by 𝑥 to the power of two-fifths. Therefore, 𝑛 is two-fifths. So we add eight multiplied by 𝑥 to the power of 𝑛 plus one which is 𝑥 to the power of seven-fifths. And we then divide by seven-fifths. And we mustn’t forget to add our constant of integration 𝐶. Now, all that remains to do is to simplify. And so we obtain a solution that the indefinite integral of negative four multiplied by the fifth root of 𝑥 to the power of nine plus eight all multiplied by the fifth root of 𝑥 squared with respect to 𝑥 is equal to negative five multiplied by 𝑥 to the power of sixteen-fifths over four plus 40 multiplied by 𝑥 to the power of seven-fifths over seven plus 𝐶.

We’ve now covered a variety of examples of indefinite integrals of power functions. Let’s recap some key points of the video. Key points. The indefinite integral of lowercase 𝑓 of 𝑥 with respect to 𝑥 is equal to capital 𝐹 of 𝑥 plus 𝐶 where capital 𝐹 of 𝑥 is the antiderivative of lowercase 𝑓 of 𝑥 and 𝐶 is a constant. The power rule for integration. The indefinite integral of 𝑥 to the power of 𝑛 with respect to 𝑥 is equal to 𝑥 to the power of 𝑛 plus one over 𝑛 plus one plus 𝐶 for any real number 𝑛 except negative one.