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Class 12 math (India)
Course: Class 12 math (India) > Unit 10
Lesson 14: Definite integral as the limit of a Riemann sumRanking area estimates
A good way to test your understanding of Riemann sums is to rank the values of various different sums.
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At 3.33 you move on to the next expression, but you use right handed rectangles, what determined that it was left or right? The only difference in the expression was the starting point of N....It is stated that because it is multiplied by one, but so is the previous expression.
Thanks!(18 votes)
Right and left handed rectangles can be determined by looking at the index in the summation. Notice that the first example starts at i=0. This means that we start the Riemann sum at x= -5, and move forward in 1 unit increments to the right. This is obviously a left handed sum. Now look at example 2. We start the index at i=1, which means the first x value evaluated in f(x) is x= -4. This means that this is a right handed sum. If you ever have issues determining which side the rectangles are based on, try the first few values in the summation.(29 votes)
- why is the definite integral equal to the exact area under the curve ? isnt it the sum of infinitely small rectangles ?Aren't we either way giving up infinitely small piecesof area?(5 votes)
As you just say, we are either overestimating or underestimating by an "infinitely small" margin, that means that the error is infinitely small, so for all practical purposes the error is 0 and the area under the curve is exact.(10 votes)
- are right handed rectangles always overestimate and left handed underestimate?(4 votes)
No. it depends on the shape of the curve. Look at the estimates for f(x) = e^(-x)(10 votes)
How did you know the first expression wanted left-handed rectangles vs right-handed rectangles?(2 votes)
Because the first one, the "i" starts with 0, which means the first height we will consider is f(-5). f(-5) is on the left side of the rectangle, I hope you can see it.
On the other hand, on the second expression, "i" starts with 1, which means our first height is f(-5+1) which equals to f(-4). f(-4), as you can see, is the right side of the rectangle.
Furthermore, when "i" equals 10 on the second expression, our height is f(5), the right side height of our last rectangle. Similarly, on the first expression, when "i" equals 9, we consider the height as f(4), the left side. Hope you understand it.(5 votes)
- So, the video is about ranking area estimates, and at the end he ranks them from largest to smallest, but he never ranks them by accuracy. Obviously the definite integral is the most accurate, and the i --> 20 is the next most accurate, but is there any general rule, perhaps based on concavity that allows us to judge left vs right handed rectangles in terms of accuracy?(3 votes)
As you noticed, accuracy is more a function of subdivisions than right handed or left handed rectangles. Since functions can be unpredictably curvy, one rule may be more accurate than another (then assuming the same number of subdivisions). Only the definite integral is guaranteed to be exact all the time.(2 votes)
Does whether it is a right handed or left handed rectangle depend on the equation? Could the same expression that you would consider left handed rectangles on one graph be right handed rectangles on another? Thanks, and will vote up an acceptable answer.(2 votes)
In this video, Sal is trying to find an estimate of the blue area. If you see the first summation, you will need to consider the values f(-5), f(-4),...,f(4). If you find the right handed area at x= -5 i.e. f(-5)*delta(x) gives the area of the rectangle right of x= -5, then you would be calculating outside of the blue area. Therefore, left-estimate was better choice here.
Similarly for the second summation, if you find the left handed area at x=4 i.e. f(4)*delta(x) gives the area of the rectangle right of x=4, then you would again be calculating outside of the blue area. So right handed rectangles are used.
Hope this helps!(2 votes)
Couldn't we just take the average of the right-handed and the left-handed to make the estimated area way more accurate? Just thinking.(2 votes)
A lot of times this does help, but it most cases it is still inaccurate. Imagine trying to do this with a very tight sine wave, or a very zig-zaggy curve. The estimates might come out very different. Sometimes the right-handed estimate will be more positive then the left-handed estimate is negative, and then you will get slightly inaccurate results.(2 votes)
- what handed is the last equation actually? right handed I guess? because for it to be left-handed, It needs to start right from -5 which can be done only by saying from 0 to infinite. But when we expand the last equation is it from 0 to infinite or 1 to infinite?(2 votes)
It's neither, since the area is infinitely small it doesn't matter which side you start at.(1 vote)
Wouldn't the Riemann Sum of the third equation approach infinity as the upper-bound of the sum approached infinity? You can see that if n=100, then f(45)(1/2) isn't even on the graph!(1 vote)- With the Riemann Sums the increment between f(x)'s, that is the space between i's gets infinitely small, so for n = 100 it would be f(-5 + i/10)(1/10), for n = 1000 it would be f(-5 + i/100)(1/100) ...so you stay within the bounds of [-5,5]. The space between rectangles(dx) --> 0 (kind of) and n --> infinity.(2 votes)
- When i= 1 it doesnt always give us a right hand-handed rectangle right? It also depends on how we construct the function I assume.(1 vote)
Video transcript
- [Voiceover] Right over
here we have the graph of F and then we have four
different expressions. What I encourage you to
do is pause this video and see if you can
figure out which of these expressions would give
the largest quantity, the second largest quantity, the second smallest quantity and
the smallest quantity. I'm assuming you have paused the video and you have given an attempt. Now let's work through this together. This first expression right over here, we're taking the sum from
I equals zero to nine. We're actually taking
the sum of 10 things, because we're taking
the zero thing, first, second, third, all the way up to nine. So this is actually going
to be the sum of 10 things because we're starting at zero. We start at F of negative five plus zero. We're saying negative five, F of negative five plus zero. That's this height right over here. That's that height right over there. We're going to take that,
and then when I equals one it's going to be negative five plus one, which is negative four. It's that height right over there. Then negative five plus
two when I equals two. It's going to be that
height right over there. We're essentially going
to sum up all the way. This is going to be negative five, negative five all the way up to, negative five plus nine is going to
get us all the way to four. It's going to be all the way over there. All the way over there. You might be guessing,
"Well, how do I relate this? "They've already kind of made us think "that we're going to somehow relate this "to area somehow, but how do we "actually make that relationship?" Because right now, as this is written, it's just giving us essentially a bunch of the values of the
functions at different points. I guess you could say it's a lot of these heights, right over here. But one thing that might jump out at you is you could construct
rectangles, all that have width one, and so if
you multiply the height times the width, the area is going to be the same thing as the height. If we put a one times one right over here, this makes it very
clear that you're taking the height times the
width of this rectangle and then this rectangle. You essentially have
a bunch of left-handed rectangles that you could imagine are trying to estimate this bluish
area that was shaded in. It's clearly going to be an underestimate, because it's giving up these areas. It's giving up those
areas, right over there. All of these rectangles are sitting, they're either just touching or they are below the actual function. Let me just write this right over here. This is going to be an underestimate of the area of this blue area. Now let's think about
what this one is, here. This is F, it's the same thing
that we're taking the sum of. We're starting at I equals
one and we're going to 10. Once again, 10 things. Negative five plus one is negative four. F of that is this line, right over here. Is that line right over there. It looks like we're taking
right-handed rectangles, because you could say times one, times one would be the area. Obviously, if we multiply
by one, we're not changing the value of this expression. That would be the area of this
first right-handed rectangle. Then when I is equal to
two, it's going to be F of negative three, and so you're looking at this one right over here. I can, let me draw at
least this part of it. It's going to look something like this. It's going to look something like this, where now we're dealing with right-handed rectangles, and these are
clearly an overestimate. These are right-handed rectangles. You're going to go all the
way, when I is equal to 10, negative five plus 10 is five F of five. That's this line right over here, or this length right over here, F of five. Of course, we're multiplying it by one. It's going to look like that,
and we could keep going. I think you get the general idea now. These are all going to be right-handed or these are all right handed
rectangles that I've drawn. These are going to be an
overestimate of the area, because they all have this little extra, they all have this extra
region, right over here. These are going to be an overestimate. Now let's think about
this one, right over here. This is, we're going to
start at I equals one and we're going to go to 20. It looks like we're going to do rectangles instead of width one, we're
going to width one half. Once again, since we're
starting at I equals one, these look like right-handed
rectangles again. This is, that we're
going to use to estimate. Let me do this in a color
that you can actually see. I'll do this in orange. I'll do this in orange right over here. The first one is going to be
negative five plus one half. It's going to be this, F
of that, which is going to take you right over there. Then you're going to multiply it times the width, which is one half. Now we have twice as many, twice as many right-handed rectangles. It's going to look like this. Twice as, and I won't do all of them because it takes some time. We're going to have twice as many right-handed rectangles. It's still going to be an overestimate, but it's going to be
less of an overestimate than this one over here. Because this one over here, you had all of this extra green
space above the function. Now we have a lot less extra space, a lot less extra space above the function. It's a better estimate, but it's still going to be an overestimate,
because at least for this function right over here, because at least over this interval where the function is increasing, the right-handed rectangle
is giving us an overestimate. But this is a little bit more precise because we're using narrower
rectangles to estimate. Overestimate but less so, but less so than this one right over here. This right over here, this
is a definite interval, from negative five to five of F of X, DX. You can imagine, this
is essentially the limit as we take these widths to be smaller and smaller and smaller. We essentially end up having
essentially an infinite number. We approach an infinite number of these right over here. This is the actual area. This is what's actually depicted in blue. This is the actual area. If I were to list this
from largest to smallest. The biggest overestimate, this is the biggest overestimate,
right over there. This is still going to be an overestimate but it's a little bit more precise because we have more rectangles. I'll put that two. This is the actual area. I'll put that three. Then this is actually an underestimate. I would rank this four. This is the largest of the values, and this is going to be
the smallest of the values.