Riemann Sums Right Endpoints Example 1

Hi everyone! Welcome back to integralcalc.com. Today we’re going to be doing another Riemann
Sum example. And this one, we’re going to be evaluating the function f of x equals two
minus x squared using right endpoints, we’ve been asked to evaluate on the range x equals
zero to x equals two so everything between the line x equals zero and x equals two, and
the problem has asked us to divide that region between zero and two into four sub intervals
or four rectangles so n equals four. We’re going to need to use a couple of formulas
with this problem. One is the area formula that helps us find the area between the graph
and the x axis of this function on this range with four sub intervals. And, this other formula
helps us calculate delta x which will be… will determine the width of the rectangles
or the width of the sub intervals… of the four sub intervals that we’re going to use
to find the area. So, let’s go ahead first and take a look
at the graph of this function. Our purple line here is the graph of two minus x squared
and our orange lines here are the lines x equals zero and x equals two, so, just to
give us kind of a visual representation. What we’ve been asked to do is find the area
between the… the graph of our function, the purple line, and the x axis. So we’ve
been asked to find the area bounded here by the line and the x axis to the right of the
line x equals zero, so everything here, and then everything to the left of the line x
equals two between the graph and the x axis, so these two regions right here, and we’re
going to do it with right endpoints. So, the first thing that we do with any of
these problems is calculate delta x or the width of the rectangles we’re going to be
using… or the width of our sub intervals and we’re going to use this formula to do
it. So, keep in mind that a and b, our lower and upper limits, respectively. So, because
they’ve already told us we’re going to be evaluating between zero and two that means
that zero is a and two is b. So we plug in here two for b and zero for a. And then, n
is the number of rectangles into which we divide the region or the number of sub intervals.
So, we set n equal to four and we put that in to our delta x formula. Of course, when
we simplify this, we’ll get two over four which reduces to one half, so, one half is
equal to delta x. In other words, the width of the rectangles into which we’re going
to divide this region is width one half. So, what that looks like when we kind of shading
our rectangles? Our first rectangle will be here from zero to one half, right, because
we start at zero, that’s our lower limit, and delta x is one half so we go to one half
and then one half to one. And so, the rectangles represent the width delta x all the way up
to where it meets the graph, where it meets our function.
So, let’s go ahead and now start calculating the… the rectangle regions. So, the first
thing that we’re going to do is specify the left and right endpoints of each of our
rectangles. So remember we said we’re starting at zero, because delta x is equals to one
half that means that the first rectangle starts at zero and ends at one half. The second rectangle
therefore starts at one half and ends at one, and then one to three halves, and then three
halves to two. And we stop at two because that is our upper limit of our range and it
makes sense because we have n equals four and you can see we ended up with four sub
intervals when we went one half at a time. So what that means, since we are using right
endpoints to evaluate the area is that we’re going to use the right endpoint of each of
these rectangles or the number here on the right hand side.
So, for this first rectangle we’re going to use one half, then one, then three halves,
and then two, as the points where our rectangles meet the graphs. So you can see, we’re using
these four right endpoints, the right endpoint of our range… of our rectangle, zero to
one half, is one half and so you can see that we’ve shaded in the rectangle up to one
half. Then, for the rectangle that goes from one half to one, we’re using the right endpoint
which is one, that’s the right side of the rectangle one half to one and this pink point
here is where it meets the graph. Same thing with one to three halves, the right endpoint
is three halves and at three halves the graph is right here so we have our pink dot. And
again, here for three halves to two, this is the right end point of this rectangle,
and that’s why we pinked the rectangles that we did.
So, our Riemann sum representation of this area is R sub four because n is equal to four,
we’re using four sub intervals. i is always equal to one. We have four up here because
we’re going to be using four points of evaluation or four rectangles or sub intervals. f of
x sub i represents the sum of our function, we’ll be using the area formula up here
to calculate that, and then that’s multiplied by delta x which, you remember we already
calculated, was one half. So, let’s use these four right endpoints
to plug in to our formula. What that looks like is the following. Remember we use that
area formula and the area formula said that we were going to have delta x out in front
here and we’ve calculated the delta x with one half, and then, that we were going to
plug in our four right endpoints which were one half, one, three halves, and two which
we have here and we’re going to plug those in to our original function f of x so we’re
going to plug it in at the function two minus x squared and evaluate at those points, that
will calculate the area of those green rectangles and give us the area between the graph and
the x axis using right endpoints. So when we simplify this, we’ll plug in one half
to our original function and that gives use two minus one half squared so you see that’s
the first term here. When we plug in one, we’ll get two minus one squared so you can
see the second term. We plug in three halves to the original function then we plug in two
to the original function. And now, to simplify this, this first term here, we get one half
squared which is one fourth, two minus one fourth is seven fourths so seven fourths is
the result there, two minus one gives us the one here, two minus three halves squared gives
us a negative one fourth so we end up with negative one fourth, and then a minus two
from this term. When we find a common denominator within the parenthesis, we’ll put everything
over four. And then, when we combine everything here inside the parenthesis, seven fourths
plus four fourths minus one fourth minus eight fourths is two fourths. And, of course, we
still have everything multiplied by delta x which is one half so when we now multiply
these two together, the two’s will cancel and our final answer for the area is one fourth.
So, one fourth is the area between the graph and the x axis on the range zero to two when
we divide that area into four sub intervals or use four rectangles to calculate and we
use the right endpoints of those rectangles. Because this answer is positive, it also tells
us that there was more area between the graph and the x axis above the x axis than below
it. If the number were negative, it would said that there was more area beneath the
x axis than there was above. So that’s it. I hope it helped you guys
and see I’ll see you in the next video. Bye!