Question Video: Finding the Integration of a Polynomial Function Using the Power Rule | Nagwa Question Video: Finding the Integration of a Polynomial Function Using the Power Rule | Nagwa

# Question Video: Finding the Integration of a Polynomial Function Using the Power Rule Mathematics • Second Year of Secondary School

Determine β«(25π₯Β² β 65π₯ + 36) dπ₯.

06:38

### Video Transcript

Determine the indefinite integral of 25π₯ squared minus 65π₯ plus 36 with respect to π₯.

All this notation means is that weβre looking for a function whose derivative is the highlighted expression 25π₯ squared minus 65π₯ plus 36. The first thing to note is that we can find the antiderivative of each term individually. Why is that? Well, if we find a function that differentiates to 25π₯ squared β in other words, the antiderivative of 25π₯ squared β and the antiderivatives of the other terms too β in other words, two other functions whose derivatives are 65π₯ and 36, respectively β then if we differentiate this combination of the functions π of π₯ minus π of π₯ plus β of π₯, then using one of the properties of differentiation, that the derivative of a sum or difference is equal to the sum or difference of the derivatives, we get 25π₯ squared minus 65π₯ plus 36.

You might like to pause the video here and take a few seconds to think about this. And this quadratic expression here is what we want to find the antiderivative of. So weβve taken our problem to find what differentiates to 25π₯ squared minus 65π₯ plus 36. And weβve shown that we can solve this if only we can find the antiderivatives of 25π₯ squared 65π₯ and 36, individually. Weβve taken one big problem and turned it into three smaller problems.

So letβs try to find functions π, π, and β with the required properties. Notice, at each of these derivatives is a monomial. That is, they all have the form π times π₯ to the power of π, where π is a whole number. So if we can find the antiderivative of something of this form, then weβre done. Can you think of a function whose derivative is 36? Pause the video to think about this.

The slope of this functionβs graph is constant. Itβs always 36, no matter where you are. And so this must be a straight line graph. The function β of π₯ could therefore be 36π₯. This functionβs graph is a straight line with slope 36. And indeed, the derivative of the function is 36, if you work it out. Of course, the function β of π₯ equals 36π₯ plus one would work just as well, as would 36π₯ minus seven. In fact, any constant would work here. Weβll come back to this point later. But for now, weβll try to find a function π of π₯ which differentiates to 65π₯.

Remember that when finding the derivative of a power of π₯, you reduce the exponents by one. The derivative of π₯ to the π is π times π₯ to the π minus one. And more generally, we can find the derivative of a constant multiple of a power of π₯. The derivative of π times π₯ to the π is π times ππ₯ to the π minus one. Now, if we compare this right-hand side, π times ππ₯ to the π minus one, to 65π₯, we can see that π times π must be 65. From the coefficient of this term and by looking at the exponent, we can see that π minus one must be one. Itβs then not hard to solve these two equations. π must be two. And using this value of π in the other equation, we see that π must be 65 over two. So what differentiates to give 65π₯? Using the values of π and π that we found, we see the answer is 65 over two π₯ squared. And you can check, if youβd like, that this really does differentiate to give 65π₯.

Our final task is to find something that differentiates to 25π₯ squared. We compare the 25π₯ squared to π times ππ₯ to the π minus one. And we see that π times π must be 25 and π minus one equals two. Solving these again is straightforward. π is three and π is 25 over three. So π of π₯, which is aπ₯ to the π, is 25 over three π₯ cubed. We found π of π₯, π of π₯, and β of π₯. Remember that we showed that the antiderivative we were looking for can be written in terms of these π, π, and β. And we can substitute in the expressions we found for them. Is this our answer then? Well, not quite.

Although you can check that this does indeed differentiates to give 25π₯ squared minus 65π₯ plus 36, there are other functions which do too. For any constant πΆ, whether itβs one or negative seven or some other number, 25 over three π₯ cubed minus 65 over two π₯ squared plus 36π₯ plus πΆ will also differentiates to give the required quadratic expression. This is the antiderivative of 25π₯ squared minus 65π₯ plus 36. Or, alternatively, using different terminology, it is the indefinite integral of this quadratic expression. You mustnβt forget the plus πΆ.

We answered this question by using what we knew about differentiation. For example, at the start of the video, we showed, using the properties of differentiation, that we could find the antiderivative of each term individually and then combine them. And we also used what we knew about the derivative of a monomial function, π times π₯ to the π. And we found that by using this formula, we could set up some simultaneous equations from the coefficient and exponent, that we could then solve to find the values of π and π that we needed.

But of course, it would have been more convenient had we known whatβs to differentiate in order to get π times π₯ to the π. Then we could have avoided solving an equation each time. It turns out that the thing we needed to differentiate to get π times π₯ to the π is π over π plus one times π₯ to the π plus one. This formula works as long as π is not equal to negative one. If π is negative one, then we have zero in the denominator which is a problem. We can also express this last formula using indefinite integral notation, where again itβs important not to forget the plus πΆ.

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