Video Transcript
Determine the integral of three 𝑥 to the ninth power minus eight 𝑥 to the eighth power minus five 𝑥 to the fifth power minus two with respect to 𝑥.
In this question, we’re asked to integrate a function with respect to 𝑥. And we can see that this is a polynomial function. And we recall that we can evaluate the integral of a polynomial function term by term by using the power rule for integration. The power rule for integration tells us for any real constants 𝑎 and 𝑛, where 𝑛 is not equal to negative one, the integral of 𝑎𝑥 to the 𝑛th power with respect to 𝑥 is equal to 𝑎 multiplied by 𝑥 to the power of 𝑛 plus one divided by 𝑛 plus one plus the constant of integration 𝐶. We add one to our exponent of 𝑥 and then divide by this new exponent.
We want to apply this term by term to integrate our polynomial function. Let’s start with the leading term three 𝑥 to the ninth power. We want to add one to the exponent of 𝑥, giving us a new exponent of 10, and then divide by this new exponent. This gives us three 𝑥 to the 10th power divided by 10. And we could add a constant of integration when we integrate each term. However, we would then just combine this into one constant of integration at the end.
So let’s just integrate our next term. We need to integrate negative eight 𝑥 to the eighth power. We need to add one to the exponent to give us a new exponent of nine. And then we divide by this new exponent. So we need to subtract eight 𝑥 to the ninth power divided by nine. We can then do the same for our next term. We add one to our exponent of five to get a new exponent of six and divide by this value. We get negative five 𝑥 to the sixth power divided by six.
There’s a few different methods of finding the integral of the constant term in our polynomial. One way would be to write this constant as negative two times 𝑥 to the zeroth power and then apply the power rule for integration. We would add one onto our exponent of zero to get a new exponent of one and then divide by this new exponent, giving us negative two 𝑥 to the first power all divided by one, which we could simplify to give us negative two 𝑥. However, it’s usually easier to remember that we can integrate any constant by multiplying it by 𝑥. And this is because it gives us an antiderivative of that function. For example, the derivative of negative two 𝑥 with respect to 𝑥 is negative two. In either case, we need to subtract two 𝑥. And remember, we need to add our constant of integration 𝐶. And since there are no shared factors in the numerator and denominator, we can’t simplify this expression any further.
Therefore, we were able to show the integral of three 𝑥 to the ninth power minus eight 𝑥 to the eighth power minus five 𝑥 to the fifth power minus two with respect to 𝑥 is equal to three 𝑥 to the 10th power over 10 minus eight 𝑥 to the ninth power over nine minus five 𝑥 to the sixth power over six minus two 𝑥 plus a constant of integration 𝐶.
