Video Transcript
Given 𝑓 of 𝑥 is equal to 𝑥
squared minus four and 𝑥 is greater than or equal to negative four and less than or
equal to two, evaluate the Riemann sum for 𝑓 with six subintervals, taking sample
points to be midpoints.
Remember, we can approximate the
definite integral of a function between the limits 𝑎 and 𝑏 by splitting the area
between the curve and the 𝑥-axis into 𝑛 subintervals. The width of each rectangle is
given by 𝛥𝑥, which is found by subtracting 𝑎 from 𝑏 and dividing this value by
𝑛. In this example, we want to split
the area into six subintervals. So we let 𝑛 be equal to six, and
then 𝑎 is negative four and 𝑏 is two. The width of each of our
subintervals is given then by two minus negative four over six, which is equal to
one. Next, we’re going to sketch the
curve of 𝑦 equals 𝑥 squared minus four between the endpoints negative four and
two. And we’re going to split this into
six subintervals.
This question requires us to use
midpoints. So the height of each rectangle
will be equal to the value of the function at the middle of each subinterval. That’s going to look a little
something like this. Now, we can work out the 𝑥-value
at the end of each rectangle by repeatedly adding one to four. And that gives us each of these
values. We can then see that the midpoints
are going to be negative 3.5, 𝑥 equals negative 2.5, negative 1.5, and so on.
Now, we’re going to need to be
extra careful here. We see that some of our rectangles
sit below the 𝑥-axis. This means we’re going to find the
negative value of their area. In other words, we’ll subtract the
total area of the rectangles that sit below the 𝑥-axis from the total area of the
rectangles that sit above the 𝑥-axis. Let’s begin by finding the function
values at each midpoint. And that will tell us the height of
each rectangle. That’s 𝑓 of negative 3.5, 𝑓 of
negative 2.5, 𝑓 of negative 1.5, and so on. And, of course, our function is 𝑥
squared minus four. And when we substitute each of
these values into that function, we get 8.25, 2.25, negative 1.75, negative 3.75,
another negative 3.75, and another negative 1.75. We then obtain the area of our
first rectangle here to be 8.25 times one.
Our second rectangle has an area of
2.25 times one. Our third rectangle has an area of
1.75 times one, not negative 1.75. Because we’re just dealing with
areas at the moment. Our fourth rectangle has an area of
3.75 times one. Base times height with our fifth
rectangle is 3.75 times one again. And for our last rectangle, it’s
1.75 times one. The Riemann sum is therefore 8.25
plus 2.25 minus the sum of 1.75, 3.75, 3.75, and another 1.75. And that gives us an approximation
to the definite integral between the values of negative four and two of 𝑥 squared
minus four. It’s negative 0.5. And, of course, 𝑥 squared minus
four is a fairly simple function to integrate. So we can check our answer by
evaluating that integral.
When we do, we get 𝑥 cubed over
three minus four 𝑥 between the limits of negative four and two, which gives us a
value of zero. And our estimate of negative 0.5 is
pretty close. So we can assume we’ve probably
done this correctly. It’s worth noting that a sketch of
the curve won’t always be possible. So instead, we need to notice that
when the value of 𝑓 of 𝑥 is less than zero, we subtract the area of the rectangle
with that height.
