### 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.