### Video Transcript

Determine the integral of two π to the nine π₯ minus π₯ with respect to π₯ between
the limits of one and zero.

Weβre actually going to be integrating two different functions here. Weβre going to integrate two π to the power of nine π₯ with respect to π₯ and
negative π₯ with respect to π₯. Since negative π₯ is the same as negative one π₯, we can split this integral up. Itβs the same as the integral between the limits of one and zero of two π to the
nine π₯ with respect to π₯ minus one times the integral between the limits of one
and zero of π₯ with respect to π₯.

And in fact, we can even simplify this first integral a little bit by taking out a
factor of two. Weβre allowed to do this. It just means that at the end of our calculation, weβre going to multiply everything
by two rather than doing it at the beginning. So, how do we integrate π to the power of nine π₯?

Well, we can think of integration as the opposite of differentiation. And we know that when we differentiate π to the power of π₯ with respect to π₯, we
simply get π to the power of π₯. So, when we integrate π to the power of π₯ with respect to π₯, we also get π of the
power of π₯ plus that constant of integration π. Well, we can extend this and say that if we differentiate π to the power of some
constant π multiplied by π₯, we get π times π to the ππ₯. And this means the integral of π to the ππ₯ with respect to π₯ is one over π π to
the ππ₯ plus that constant of integration π.

That means the integral of π to the power of nine π₯ with respect to π₯ is a ninth
π to the power of nine π₯. So, two times the integral of π to the power of nine π₯ is two times a ninth π to
the power of nine π₯. But weβre actually gonna evaluate a ninth π to the power of nine π₯ between the
limits of one and zero before multiplying by two. The integral of π₯ is a little bit more straightforward. Itβs π₯ squared divided by two.

Our next step is to evaluate each of these expressions between one and zero. Evaluating a ninth π to the power of nine π₯ between the limits one and zero, and we
get a ninth π to the nine times one minus a ninth π to the nine times zero. And if we evaluate π₯ squared divided by two between one and zero, we get one squared
divided by two minus zero squared divided by two. Zero squared divided by two is simply zero. And nine multiplied by zero is also zero. So, we have a ninth π to the power of zero. And since π to the power of zero is one, this just becomes one-ninth.

And simplifying, we see that we have two times a ninth π to the power of nine minus
a ninth all minus one-half. Weβll distribute this first set of parentheses by multiplying each term one-ninth π
to the nine and negative a ninth by two. And our final step is going to be to evaluate negative two-ninths minus a half.

To evaluate that, weβll need to make the denominators the same. The lowest common denominator of nine and two is 18. So, we need to multiply both the numerator and the denominator of two-ninths by two,
and the numerator and the denominator of one-half by nine. Two-ninths is, therefore, equivalent to four eighteenths, and a half is equivalent to
nine-eighteenths. Negative four eighteenths minus nine eighteenths is negative thirteen
eighteenths. So, the integral of two π to the nine π₯ minus π₯ with respect to π₯ between the
limits of one and zero is two-ninths π to the power of nine minus thirteen
eighteenths.