### Video Transcript

Evaluate the definite integral
between negative four and two of negative π₯ minus four with respect to π₯ using the
limit of Riemann sums.

Remember, if π is some integrable
function on the closed interval π to π, then the definite integral between π and
π of π of π₯ with respect to π₯ is defined as the limit as π approaches β
of the sum of π of π₯π times Ξπ₯ for values of π from one to π. And of course, Ξπ₯ is π minus π
over π and π₯π is π plus π lots of Ξπ₯.

Letβs begin by comparing this
definition with our integral. We see that π of π₯ is equal to
negative π₯ minus four. This is a polynomial. And We know polynomials are
continuous over their domain. So the function negative π₯ minus
four is continuous and therefore integrable over the closed interval defined by the
lower limit negative four and the upper limit two.

We therefore let π be equal to
negative four and π be equal to two. And weβll next look to define
Ξπ₯. Itβs π minus π. So thatβs two minus negative four
over π. That gives us six over π. Next, weβll define π₯π. Itβs π, which is negative four,
plus π lots of Ξπ₯, which we calculated to be six over π. This gives us that π₯π is negative
four plus six π over π.

It follows that we can find π of
π₯π by substituting this expression into our function. That gives us negative negative
four plus six π over π minus four. When we distribute the parentheses,
we find π of π₯π to be equal to four minus six π over π minus four. And of course, four minus four is
zero. So π of π₯π is simply negative
six π over π.

And we can now substitute
everything we have into our definition for the integral. This tells us that the definite
integral between negative four and two of negative π₯ minus four with respect to π₯
is equal to the limit as π approaches β of the sum of negative six π over
π times six over π for values of π from one to π. Now in fact, negative six π over
π times six over π is negative 36π over π squared. So this is the sum.

Now of course, this factor,
negative 36 over π squared, is actually independent of π. So we can rewrite our limit as the
limit as π approaches β of negative 36 over π squared times the sum of π
from values of π from one to π. And in fact, whilst itβs outside
the scope of this video to prove this, we can quote the general result. The sum of π from π equals one to
π is equal to π times π plus one over two. And this means we can replace all
of this sum with the expression π times π plus one over two.

Our definite integral can therefore
be evaluated by working out the limit as π approaches β of negative 36 over
π squared times π times π plus one over two. And you might spot that we can
divide both 36 and two by two. We can also cancel one π. And our limit reduces to the limit
as π approaches β of negative 18 times π plus one over π.

Letβs distribute our
parentheses. And when we do, we find that this
can be written as negative 18π over π minus 18 over π. Now of course, negative 18π over
π is just negative 18. And we can now evaluate our limit
by direct substitution. As π approaches β, negative
18 over π approaches zero. And so the limit as π approaches
β of negative 18 minus 18 over π is simply negative 18. We can therefore say that the
definite integral between negative four and two of negative π₯ minus four with
respect to π₯ is negative 18.