Show that, in general, RN-LN=(b-a) ×(f(b)-f(a))/(N) |...

Name: Problem 55, Chapter 5: Calculus Volume 1
Uploaded: 2020-03-25T18:12:08-08:00
Duration: 11 min 50 s
Channel: Darren Mckaig
Description: Show that, in general, RN-LN=(b-a) ×(f(b)-f(a))/(N)

Key Concepts

Riemann Sums

Riemann sums are a fundamental technique in calculus used to approximate the area under a curve by dividing the region into a finite number of subintervals and summing the areas of simple figures, typically rectangles, whose heights are determined by the function values at specific sample points. This method serves as the basis for defining the definite integral as the limit of these sums as the number of subintervals increases indefinitely.

Left and Right Riemann Sums

Left and right Riemann sums are specific types of Riemann sums where the sample point in each subinterval is chosen to be the left endpoint or the right endpoint, respectively. The use of different endpoints results in slightly different approximations of the integral. Analyzing the difference between these two sums can provide insights into the behavior of the function over the interval, particularly regarding whether it is increasing or decreasing.

Partition of an Interval

A partition of an interval involves dividing the interval [a, b] into a number of subintervals, often of equal width, where the width is calculated as (b - a)/N for N subintervals. This partition serves as the framework for constructing Riemann sums, as it determines the size of each rectangle and, consequently, affects the accuracy of the area approximation under the curve.

Telescoping Sums

Telescoping sums are series where successive terms cancel out a portion of preceding or following terms, allowing the series to collapse into a simpler form. In the context of comparing left and right Riemann sums, the difference between the two can often be rearranged into a telescoping sum, which then simplifies to an expression that directly relates to the change in the function values over the interval.

Transcript

00:01 Okay, for problem 55, we're going to try to show that the right -hand riemann sum minus the left riemensum is equal to this product over here.

00:16 And this is going to be pretty complicated.

00:20 First, we're going to write out the formula for how we use a summation to find a right -hand ream and it's going to stay all conceptual.

00:31 So i'll try to explain these ideas as i go, but first of all, we have b minus a, which is the length of the entire interval divided by n, which is the number of rectangles you want.

00:51 So this would be, this represents the width of one rectangle.

00:58 And then we're going to sum this.

01:04 Let's go ahead and erase that, move it down so i can write on top of it.

01:18 So let's go back to red here.

01:24 There's always a certain amount of rectangles we're going to use in each remand sum.

01:31 So we're going to go from the first one all the way to the last rectangle, n.

01:39 And then remember, this is the width of the rectangle out here.

01:43 And what we're summing are all the heights of all the rectangles.

01:46 The width is going to be the same, but the heights are going to change.

01:51 And so in here we have to represent the height, which is the, we find by plugging our x values into the function.

02:03 So it's going to be a function, whatever our function is, evaluated at each individual x value.

02:12 Well, our first x value is always a, because we're going from a to b.

02:19 So that is our first x value.

02:23 But then we get to our next x value by adding the width of each rectangle.

02:29 So we have to add this little fraction right here.

02:35 B minus a over n.

02:41 That's how we get to the next rectangle.

02:44 But then we have to get to each subsequent rectangle, so we'll have to keep adding this over and over and over again.

02:51 So we have to put a multiplier out here.

02:56 I.

02:58 And that's how we're going to keep moving down the line.

03:00 We'll plug in i equals one to move over one rectangle.

03:04 I equals two to move over two rectangles.

03:07 And so plugging all of this in the original x value plus the width of each rectangle to get to each subsequent x value plugged in to the function will give us all of our heights or y values.

03:23 And then times the width, width times height will give us the area of the rectangle.

03:29 So that's how, why that's the formula for finding the right hand ream and sum.

03:36 Now, we'll explain the difference here with the left hand ream and sum.

03:40 It's slightly different, but most of it's the same.

03:46 With the rectangle times the summation of all the heights.

03:58 Still going to need to find our heights using the function.

04:06 Okay, the difference here, you're going to see in a second, is when i plug i equals 1 in, it means i'm moving over one x value before i even evaluate my first rectangle.

04:27 And that's what we do when we're talking about the right -hand side of a rectangle.

04:34 Say i have a function here, and there's my, here's my first x value, x1, there's my next x value.

04:50 Well, if i'm doing a right -hand ream and sum, i want to use this side.

04:55 So i want to take my first x -value plus the width to get to this other x value before i even evaluate my first rectangle.

05:05 So the first x value plus the width.

05:08 So this is perfect.

05:11 But here, we actually have to, for the left -hand remand sum, we want to evaluate this first height by just using the first x value, just using f of a.

05:25 So really, i don't even want to add anything to find my first height of the left -hand remand sum.

05:34 So i need to put, i need to move it back a step by putting i -minus -1.

05:39 In here.

05:42 So when i plug a 1 in, 1 minus 1 is 0, 0 times this fraction, i'm not even going to add this fraction for my very first one...

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