# Video: Finding the Integration of a Polynomial

Determine β« (π₯ β 6)(π₯ β 5)(π₯ β 3) dπ₯.

02:14

### Video Transcript

Determine the indefinite integral of π₯ minus six multiplied by π₯ minus five multiplied by π₯ minus three.

Letβs start by expanding the brackets. Expanding the first two sets of brackets, we get π₯ squared minus 11π₯ plus 30. And we then multiply this by π₯ minus three. We obtain the indefinite integral of π₯ cubed minus 14π₯ squared plus 63π₯ 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 π₯ to the power of π with respect to π₯ is equal to π₯ to the power of π plus one over π plus one plus πΆ. If instead we were integrating π₯ to the power of π multiplied by some constant π, then since we can factor a constant out of our integral, then this would simply be equal to π multiplied by π₯ to the power of π plus one over π plus one plus πΆ.

Now, you may be wondering why we havenβt multiplied the πΆ by π. And thatβs because π is also a constant. Therefore, π multiplied by πΆ is a constant. And we can just rename this new constant to be πΆ. So now, letβs apply this rule to our integral term by term. The first term is π₯ cubed. Therefore, π is equal to three. We increase the power by one and divide by the new power to get π₯ to the power of four over four. The next term is negative 14π₯ squared. Negative 14 is just a constant. So that will remain. Our power is two. So π is two. We increase the power by one to get π₯ cubed and divide by the new power.

On next term, we have 63π₯. So we can start by writing in our constant of 63. We then note that π₯ is equal to π₯ to the power of one. So π is equal to one. So when we integrate this, we get π₯ 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 π₯ to the power of zero since π₯ 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 π₯ is zero, we increase the power by one giving us π₯ to the power of one and divide by the new power. So thatβs dividing by one. And we miss then adding our constant of integration πΆ.

We can write this out a little neater for our solution which is π₯ to the power of four over four minus 14π₯ cubed over three plus 63π₯ squared over two minus 90π₯ plus πΆ.