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Let's work through a problem where we have to use a definition of the definite integral.
The directions here are to find the value of X squared plus 2, from 0 to 3, by using
the definition of the definite integral with right Riemann sums. I've written below here
the definition of the definite integral which is this part of things. The definite integral
is here, this integral sign of F of X DX from A to B. What the definite integral represents
is the net area bounded by the curve F of X on the interval from A to B. The way that's
defined is the limit of a general Riemann sum. A general Riemann sum is like the Riemann
sums we've looked at in the past except, the partition is not necessarily even, each rectangle
could be of a different width. If you have that general Riemann sum and you take the
limit as delta goes to 0, that's the limit as all your rectangle widths go to 0, then
that's going to give you the exact net area under the curve. That's how the definite integral
is defined. That would be true, that should be true for any way that you subdivide your
Riemann sum. In practice then, what we do is when we're actually calculating these we
use an easy way of subdividing the interval, we use an evenly spaced partition. What we're
going to do is, we're going to the Riemann sums we've looked at in the past. Specifically
here we're going to use a right Riemann sum, that's the formula I've written here, with
an even partition, delta X equals B minus A over N and we're going to take the limit
as N goes to infinity. Once again, that's going to give us that exact area under the
curve. Let's start and see how this works out. What we're saying then is we are interested
in this definite integral and we're saying we're going to do this by using a right Riemann sum and take the limit
as N goes to infinity. My formula for a right Riemann sum is this F of A plus K delta X
times delta X. Let's figure out what are different pieces are. I'm looking at the interval from
0 to 3. A is going to be 3 and then delta X, I'm using an evenly spaced partition for
ease of calculation so delta X is just going to be B minus A over N which is 3 minus 0
over N so, 3 over N. I'm going to keep these N's in my calculation, I'm going to leave
everything in terms of N because in the end I'm going to take the limit as N goes to infinity.
If I plug all this in to my formula, I have this limit as N goes to infinity of the sum
from K equals 1 to N of F of—actually A should be 0 sorry about that. Right? Because
our interval was from 0 to 3 so, A should be 0—so, 0 plus K delta X, that's 3 K over
N, times delta X which is 3 over N. Now I need to work on actually computing this. This
3 over N here, that's a constant with respect to the sum because I'm summing over K. I can
go ahead and pull that outside the summation, so, 3 over N sum of F of 3K over N. What this
notation means, F of 3K over N, it means take that expression, 3K over N, and plug it into
my function. Remember my function was X squared plus 2. If I actually do that step it's going
to look like this, 3 over N sum of, so I'm plugging in 3 K over N in for X, 3K over N
squared plus 2. Now we want to go ahead and expand everything out. I still have my limit.
I still have my sum. Inside the summation, this is going to be 9 K squared over N squared
plus 2. Usually it's helpful when computing these sums to break them down so, I'm going
to leave this 3 over N out front of everything then I'm going to break this sum into two
pieces. The first piece is going to be this 9 K squared over N squared and within that
piece I can actually pull the 9 over N squared out front because, once again, that's a constant
with respect to the sum, which is over K. I get this 9 over N squared sum of K squared
and then plus the sum of 2. Now, both of these sums are things that I can use my special
summation formulas on so, that's what I am going to do next. I still have my limit, I
have this 3 over N. Then I'm going to have 9 over N squared times the sum of K squared,
from K equals 1 to the N. If you recall, that formula is N times N plus 1 times 2 N plus
1 over 6 and then, plus I have the sum of 2, so the sum of a constant is that constant
times N so, plus 2 N. We've gotten rid of the sums now, ultimately, I'm going to want
to take the limit of this to do that it's going to be really helpful if I clean things
up a little bit. Notice here that I have it in this first fraction inside the brackets
I have an N in the top and an N squared in the bottom, so I can do a little simplification
there, and a 9 up here and a 6 down here. That's going to become, we still have this
3 over N out front, and then this first fraction is going to be 3 N plus 1, 2 N plus 1 over
2 N and then I have this plus 2 N. I'm going to go ahead now and distribute this 3 over
N to the terms inside the brackets. If I do that that's going to become 9, N plus 1, 2
N plus 1 over 2 N squared plus 6. Now this is a lot nicer right? Let's take a look at
the terms here and think about how we might take these limits. The 6 is no problem right?
The plus 6 that's going to be easy to take the limit of but, this fraction may be a little
trickier. If you think about plugging in infinity there I'm going to end up getting the form
infinity over infinity. There's a few different ways I could approach that. Probably the easiest
way would actually be to use L'Hopital's Rule, so, I think that's what I'm going to do. To
do that, I'm going to go, it's going to be easier if I foil out that top so that's what
I'm going to do next. This is going to be 9 times the quantity 2 N squared plus, I have
an N and a 2 N so, plus 3 N plus 1 over 2 N squared. Then, I'm going to split up this
limit because I can't use L'Hopital's Rule on 6. This first limit is the limit as N goes
to infinity of 18 N squared plus 27 N plus 9 over 2 N squared then, I have my limit of
6 which I'll deal with later. Now we see a little more, well, the form of this function
right? It definitely is infinity over infinity and the two ways you could do this, you could
divide every term by N squared and take the limit or use L'Hopital's Rule. I'm going to
use L'Hopital's Rule. Remember L'Hopital's Rule says that if you have a 0 over 0 or an
infinity over infinity limit, you can take, its equivalent to the limit if you take the
derivative of the top and bottom. In my case, it's going to equal the limit of 36 N plus
27 over 4 N and I still have my limit of 6 hanging out here. It's still an infinity over
infinity limit so I can use L'Hopital's Rule again and I get the limit as N goes to infinity
of 36 over 4. Well, now everything's constants so this just equals, 36 over 4 which is 9,
9 plus 6 which equals 15. If 15 is the answer to the definite integral, the definite integral
from 0 to 3 of X squared plus 2 DX is 15.