### Video Transcript

Determine the definite integral
from negative one to one of negative six π₯ squared minus three with respect to
π₯.

Weβre asked to evaluate the
definite integral of a quadratic polynomial. And we know how to evaluate
definite integrals by using the fundamental theorem of calculus. So the first thing weβll do is
recall the fundamental theorem of calculus. In fact, weβll only recall the part
which helps us evaluate definite integrals. We know if our integrand lowercase
π is continuous on the interval of integration the closed interval from π to π
and we have an antiderivative of our integrand, so thatβs capital πΉ prime of π₯
will be equal to lowercase π of π₯, then the definite integral from π to π of
lowercase π of π₯ with respect to π₯ will be equal to capital πΉ evaluated at π
minus capital πΉ evaluated at π.

So this gives us a method of
evaluating indefinite integrals by using antiderivatives of our integrand. However, the first thing we always
need to check is, is our integrand continuous on the interval of integration? In this case, our interval of
integration will be the closed interval from negative one to one. So weβll set our value of π equal
to negative one and π equal to one. Next, our integrand lowercase π
will be negative six π₯ squared minus three. So we need to determine whether
negative six π₯ squared minus three is continuous on the closed interval from
negative one to one. If it is, we can then use the
fundamental theorem of calculus to evaluate this integral.

In this case, we see negative six
π₯ squared minus three is a quadratic polynomial. And we know all polynomials are
continuous for all real numbers. So in particular, this means that
this polynomial must be continuous on the closed interval from negative one to
one. So we can apply the fundamental
theorem of calculus to help us evaluate this indefinite integral. This means we need to find an
antiderivative of our quadratic.

We actually know several different
methods for finding antiderivatives. For example, we could try to
reverse the process of differentiation, and this would work. However, we already did this in the
general case for polynomials. This gave us the power rule for
integration. So we can find an antiderivative of
our quadratic by using the power rule for integration. On each term, we want to add one to
our exponent and then divide by this new exponent. This gives us negative two π₯ cubed
minus three π₯. And of course we add our constant
of integration πΆ.

And itβs worth reiterating what
this means. The derivative of negative two π₯
cubed minus three π₯ plus πΆ with respect to π₯ will be equal to our integrand. And this is exactly what it says
for capital πΉ of π₯ to be an antiderivative of lowercase π of π₯. And this will be an antiderivative
for any real value of the constant πΆ. So we can pick whichever value we
want. Normally, we pick zero because it
makes the expression the simplest. So weβll set capital πΉ of π₯ to be
negative two π₯ cubed minus three π₯.

Weβre now ready to apply the
fundamental theorem of calculus to evaluate our definite integral. Remember, this tells us that our
indefinite integral will be equal to our antiderivative evaluated at the upper limit
of integration minus the antiderivative evaluated at the lower limit of
integration. And we use notation to represent
this. We write our antiderivative inside
of the square brackets and then our limits of integration outside of this. And all this is is a shorthand
notation for writing capital πΉ evaluated at one minus capital πΉ evaluated at
negative one. So all thatβs left to do is
evaluate this at the limits of integration. We get negative two times one cubed
minus three times one minus negative one multiplied by negative one cubed minus
three times negative one.

And now we can start
simplifying. Inside of our first set of
parentheses, we get negative two times one cubed is negative two and negative three
times one is negative three. So this simplifies to give us
negative two minus three, which is of course equal to negative five. We can do the same to simplify our
second set of parentheses. We get negative two minus one cubed
is equal to two and negative three multiplied by negative one is equal to three. So our second set of parentheses
simplifies to give us two plus three, which is of course equal to five. But remember, we need to subtract
this. So we get negative five minus five,
which is of course equal to negative 10.

Therefore, by using the fundamental
theorem of calculus and the power rule for integration, we were able to show the
definite integral from negative one to one of negative six π₯ squared minus three
with respect to π₯ is equal to negative 10.