# uc3rmw_1005e01

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Hi there.
In this example, we're going to approximate
the area under the graph of the function
f of x.
The problem states, let's approximate
the area under the graph of f of x equals
x, from x equal 1 x equal 2 by
partitioning the interval 1, 2
into in part a, 2
subintervals of equal length
and choosing t to be the left endpoint.
In part b, 4 subintervals of equal length
and choosing t to be the left endpoint.
In part c, 6 subintervals of equal length
and choosing t to be the left endpoint.
And in part d, compare the
approximations found in parts
a through c with the actual area.
We let A be the area of a
region below the graph of y
equals x, above the x-axis, and between the
vertical lines x equals 1 and x equals 2.
In part a, we're going to find
2 subintervals of equal length
and choose t to be the left endpoint.
Here, the graph shows that
the interval from 1 to 2
is split into 2 intervals of equal length.
We have one interval from 1 to 3 over 2,
and another interval from 3 over 2 to 2.
They're both of equal length, 1/2.
So we can write that down that we take
the endpoints of our interval from 1 to 2.
We have 2 minus 1 divided by 2 equals 1/2.
That is the width of each
of these subintervals.
Therefore, the subintervals are
1, 1 plus 1/2 and 1 plus 1/2, 2.
We can simplify both of these subintervals.
We have 1, 3/2 and 3/2, 2,
with left endpoints for both
of these subintervals of 1 and 3/2.
So now, we can find the
approximate area under the curve
of f of x equals x on that
interval from 1 to 2 as follows.
The area is approximately f of 1.
That is the value of f at x equals 1
times the width of the first subinterval.
So that's 1/2.
We're going to go ahead
and add to that f of x
evaluated at the left endpoint
for the second subinterval.
That is 3/2, again, times the width
of the second subinterval, so 1/2.
So this is equal to f of 1 is 1.
We're going to multiply that by 1/2.
f of 3/2 is 3/2.
And we multiply that by 1/2.
We go ahead and simplify.
We have 1/2 plus 3/4 is
equal to 5/4, or 1.25.
So the approximate area under the graph of
f of x equals x on the interval from 1 to 2
based on 2 subintervals
with width 1/2 is 1.25.
So now, we're going to do
the same thing in part b,
but this time, we have 4
subintervals instead of 2.
Again, our interval is from 1 to 2.
And we've divided that interval
into 4 subintervals of equal width.
So to find the width of
the subintervals, we're
going to take the
difference of the endpoints.
So we have 2 minus 1 divided by
4, the number of subintervals.
And we have that the width
of the subintervals are 1/4.
Therefore, we have 4
subintervals where we start
with the left endpoint of
the first subinterval as 1.
The subintervals are 1, 1 plus 1/4;
1 plus 1/4, 1 plus 1/4 plus 1/4.
Then we have 1 plus 1/4 plus
1/4 to 1 plus 1/4 plus 1/4.
And we add to that another 1/4 for
the width of the next subinterval.
And finally, we have a fourth
subinterval with this left-most endpoint
so it's 1 plus 1/4, plus 1/4, plus 1/4.
And the right-most endpoint
to this last subinterval
is 2, adding 1/4 to this end point.
We can simplify all these subintervals down.
So we have 1, 1 plus 1/4 is 5/4, 5/4, 6/4
or 3/2, for the third subinterval 3/2.
Adding a fourth to that 3/2 gives us 7/4.
And finally, we have for the
fourth subinterval, 7/4, 2.
So now, we can find the
approximate area under the graph
of f of x equals x on
that interval from 1 to 2
using these 4 subintervals,
based on the fact
that the width of each of
these subintervals is 1/4.
The approximate area is the
evaluation of f of x, at x equals 1,
times the width of the first
subinterval, which is 1/4,
plus f of 5/4, the left-most
endpoint of the second subinterval,
times the width of that second subinterval.
We add to that f of 3/2,
again, the left-most end-point
of the third subinterval.
We multiply that by the
width of the subinterval.
And finally, we add to that
the term f of 7/4, where again,
7/4 is the left-most endpoint
of the last subinterval,
and multiply that by the
width of the subinterval.
We go ahead and evaluate f of x at
each of these left-most endpoints
to the 4 subintervals.
And we have that f of 1 is
1, and that's times 1/4.
f of 5/4 is 5/4, and that's times 1/4.
f of 3/2 is 3/2, and that's times 1/4.
And f of 7/4 is 7/4, and that's times 1/4.
All of this simplifies to 22/16, or 11/8.
And that as a decimal is 1.375.
Therefore, the approximate
area under the graph
of f of x equals x on
the interval from 1 to 2
based on 4 subintervals of
equal length, 1/4, is 1.375.
So now, we move on to part c.
We're going to evaluate the
area under the curve of f
of x equals x, on the interval from 1 to
2, using 6 subintervals of equal length.
And we're going to choose 1
to be the left-most endpoint.
So first of all, we want to find the
width of each of these 6 subintervals.
So we're going to take the
difference in the two endpoints.
We have 2 minus 1.
We're going to divide that
up by 6, and so we have 1/6
as the width of these 6 subintervals.
Therefore, we can find the left and right
endpoints to each of the 6 subintervals
as follows.
We have the left-most endpoint
for the first subinterval is 1.
We can add 1/6 to that to get 7/6.
7/6 will be the left-most
endpoint for the next subinterval.
We add 1/6 to that, and we have 4/3.
4/3 will be the left-most endpoint
for the third subinterval.
We add 1/6 to that, and we get 3/2.
3/2 is the left-most endpoint
for the fourth subinterval.
We add 1/6 to that, we have 5/3.
Next is 5/3 and we add
1/6 to that to get 11/6.
And finally, for the sixth
subinterval, we have 11/6, 2.
So now, we can approximate the
area under the curve as follows.
A is approximately f of 1 times
the width of the subinterval, 1/6,
plus f of 7/6 times 1/6, plus f of 4/3
times 1/6, plus f of 3/2 times 1/6,
plus f of 5/3 times 1/6.
And finally, f of 11/6 times 1/6.
Evaluating f of x at each of these
left-most endpoints to the 6 subintervals,
we have 1 times 1/6, plus 7/6 times
1/6, plus 4/3 times 1/6, plus 3/2 times
1/6, plus 5/3 times 1/6,
plus 11/6 times 1/6.
All this simplifies to 17/12, and that
as a decimal is approximately 1.417.
So we have that area under the curve of f
of x equals x on the interval from 1 to 2
based on 6 subintervals of equal
length, 1/6, is approximately 1.417.
So now in part d, we want to
compare the approximations found
in parts a through c to the
exact area under the curve of f
of x on the interval from 1 to 2.
We can do that using two approaches.
In the first approach,
we can actually calculate
the area of the square that's
shaded here, and then the area
of the triangle above that square.
So we have that the area
under the graph of f of x
equals x is equal to the addition
of the area of the triangle
and the area of the square.
So that the area is equal to, for
the box we have a side equal to 1.
So that's 1 times 1.
For the area of the triangle,
we have 1/2 base times height.
So that is 1/2 times 1,
the base, and the height 1.
So we can simplify that as 1 plus 1/2,
or 3/2, which as a decimal is 1.5.
So that's the first approach.
In the second approach, we can take the
area of the trapezoid formed by the area
under the curve on that
interval from 1 to 2.
That area is equal to 1/2 times 1, which
is b1, the height of the trapezoid, times,
in parentheses, the addition of the
two bases, b1 and b2 for the trapezoid.
That's 1 plus 2.
We simplify this and we have 3/2,
which again as a decimal is 1.5.
So based on these two approaches,
we have the exact area
under the curve of f of x equals x on
the interval from 1 to 2 as A equals 1.5.
So comparing our answer in part
d to our answer is in parts
a through c for the approximations of
the area under the curve on the interval
from 1 to 2, we see that as these
subintervals increase in number
from 2 to 6, that our answer gets closer
and closer to the actual area of 1.5.
More specifically, we have in part a,
the area was 1.25 for 2 subintervals.
In part b, we had 4 subintervals
and the area was 1.375.
In part c, we had 6 subintervals,
and the area got closer to 1.5.
It was 1.417.
And so the exact area of 1.5 is approached
the higher number of subintervals
that we have.