### Video Transcript

Evaluate the definite integral
between negative four and five of the absolute value of 𝑥 minus two with respect to
𝑥.

For this question, we’ve been asked
to evaluate the definite integral of a function, which we’ll call lowercase 𝑓. This function is the absolute value
or the modulus of 𝑥 minus two. Now, for any real number, we can
express an absolute value function as a piecewise function. We can do this by recalling that if
𝑥 minus two evaluates to a negative number or absolute value, we’ll multiply this
by negative one to turn it into a positive number. Okay, so when 𝑥 minus two is
greater than or equal to zero, our function is simply 𝑥 minus two. But when 𝑥 minus two is less than
zero, our function is multiplied by negative one. So it is negative 𝑥 minus two. Of course, it’s probably more
useful to us to isolate 𝑥 on one side of these inequalities. We do so by adding two to both
sides. Now, it might also be useful for us
to simplify this as negative 𝑥 plus two.

Okay. Now that we have reexpressed our
function piecewise, we can think about how it might look graphically. Here we see the graph. Although the scale is not exact,
both the graph and the piecewise definition should show us the difference in
behavior of our function either side of 𝑥 equals two. We see a sharp corner at the point
two, zero on our graph. In fact, we would say that our
function is not differentiable when 𝑥 equals two. But it is continuous when 𝑥 equals
two. This is important, because in order
to evaluate our definite integral, we’ll be using the second part of the fundamental
theorem of calculus. This allows us to evaluate a
definite integral using the antiderivative, uppercase 𝐹, of the function which
forms our integrand, lowercase 𝑓. The condition for doing so is that
lowercase 𝑓 must be continuous on the closed interval between 𝑎 and 𝑏, which are
the limits of the integration. Given that our function lowercase
𝑓 is continuous when 𝑥 is equal to two, we are able to conclude that it is
continuous over the entire set of real numbers. And hence, the continuity condition
is satisfied.

Okay, onto evaluating the definite
integral. Now we’ve already said that our
function behaves differently either side of the line 𝑥 equals two. A useful first step for us then is
to split up our integral into two parts. The first going from the lowest
bound, negative four to two, and the second going from two to five. Since the upper limit of our first
integral is the same as the lower limit of our second integral, the sum of these two
will be the same as our original integral. Now that we’ve split our integral
into two parts, we’re able to substitute in the two different subfunctions that we
defined using the piecewise definition of the absolute value of 𝑥 minus two. We can understand this by
considering our integrals as the area under these lines. From negative four to two, our
function behaves like minus 𝑥 plus two. And from two to five, our function
behaves like 𝑥 minus two. We can interpret the sum of these
two areas as being the same as our original integral.

From this point, we can now move
forward using the familiar power rule of integration. We raise the power of 𝑥 for each
of our terms and divide by the new power. Let’s tidy up to make some room for
the next steps. Here, we’ve input the limits of
both integrals. And some color has been added to
help follow the calculation. We’ll need to go through a few more
simplification steps. Again, we’ll clear some space. And we’ll continue to simplify. Eventually, we reach a point where
we’ll express everything in terms of halves. And we reach a final answer of
forty-five halves or 45 over two. With this, we’ve completed our
question. We did this by first expressing the
absolute value of 𝑥 minus two as a piecewise function. Then, by splitting our original
integral into two parts and using the second part of the fundamental theorem of
calculus to help us evaluate each individual part.