Video: Finding the Integration of a Polynomial

Determine ∫ (π‘₯ βˆ’ 6)(π‘₯ βˆ’ 5)(π‘₯ βˆ’ 3) dπ‘₯.

02:14

Video Transcript

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

Let’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 𝑛 with respect to π‘₯ is equal to π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus 𝐢. If instead we were integrating π‘₯ to the power of 𝑛 multiplied by some constant π‘Ž, then since we can factor a constant out of our integral, then this would simply be equal to π‘Ž multiplied by π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus 𝐢.

Now, you may be wondering why we haven’t multiplied the 𝐢 by π‘Ž. And that’s because π‘Ž is also a constant. Therefore, π‘Ž multiplied by 𝐢 is a constant. And we can just rename this new constant to be 𝐢. So now, let’s apply this rule to our integral term by term. The first term is π‘₯ cubed. Therefore, 𝑛 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 𝑛 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 𝑛 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 that’s dividing by one. And we miss then adding our constant of integration 𝐢.

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 𝐢.

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