Finding Area by the Limit Definition In Exercises 47-56 , use the limit process to find the area

Finding Area by the Limit Definition In Exercises 47-56 ,...

Key Concepts

Partitioning the Interval

Partitioning the interval refers to dividing the overall range of integration into smaller equal or variable segments. This process is essential for setting up the Riemann sum, where each subinterval contributes an individual area element that, when summed and taken to the limit, yields the total area under the curve.

Limit Process

The limit process in the context of integration involves taking the number of subdivisions of the interval to infinity. As the width of each subinterval tends to zero, the Riemann sum converges to the exact value of the definite integral, thereby providing an accurate calculation of the area under the curve.

Definite Integral

The definite integral of a function over an interval is defined as the limit of a Riemann sum as the number of subintervals approaches infinity. It represents the exact area under the curve (or above the curve when negative) between two bounds, and is a foundational concept in calculus for measuring accumulated quantities.

Riemann Sum

A Riemann sum approximates the area under a curve by dividing the region into small subintervals and summing the areas of rectangles or other shapes that approximate the region within each subinterval. The better the partition (i.e., the smaller and more numerous the subintervals), the closer the sum is to the actual area, leading to the limit definition of the integral.

Transcript

00:01 We're trying to use the limit process to find this area under this curve.

00:09 Okay, so the limit process is basically the remand sum.

00:18 So, i mean, you just have the integral between 0 and 1.

00:27 This is the interval of this.

00:31 Is approximately or is equal to the limit as and approaches infinity of summation f of x i times changing x okay now changing x we've been doing this already let me put this in different color changing x is b minus a over n right and this is our b and this is our a, so it's just 1 over n.

01:11 Because 1 minus 0 is just 1.

01:14 So 1 over n.

01:16 So this formula needs this and it needs l of xi.

01:22 This one too.

01:23 What is xi? xi is given as a plus changing xi.

01:32 What is our a? our a is zero in this case.

01:36 So zero plus changing x.

01:38 Is this 1 over n here, so it's just 1 over n i.

01:44 So we need this f of x i.

01:48 So f of x i, basically wherever you see x in this function, just put x i there.

01:58 So it's just going to be negative for x i.

02:02 And our x i has been shown to be this 0 plus 1 over n i.

02:11 But we do not write it 0, so it's just going to be 1 over n i.

02:15 So wherever i saw x, i put xi, which is 1 over n i there, and then plus 5.

02:24 This is what we have.

02:25 And then we need this changing x here.

02:35 So f of xi times changing x is given by negative 4 over n i plus 5.

02:48 Times changing x changing x is 1 over n so we have that now we have to take the summation of both sides so we just take summation i run from 1 to n and then summation i from 1 to n of both sides okay so this right here is going to be i'm just going to uh multiply this 1 over n by whatever you have here here so it's just going to be negative um it's going to be summation i from one to n of you know this one over n is going to i'm just distributing it to these two terms so it's going to be negative four over n squared i and then plus five over n this is what we have okay this is the summation we have now my summation properties, we can distribute the summation to each of the two terms here.

03:57 We're just going to do that on the next page.

03:59 So summation i from 1 to n of f of x, i times that is, we're not going to distribute this one to each other two terms.

04:14 So it's going to be summation i from 1 to n negative 4 over n squared i.

04:25 Can you see that? the first one, and then plus summation 5 over n.

04:34 And then furthermore, properties of summation can make us bring out constants.

04:42 Okay, so this is a constant.

04:43 It does not carry any i, so we can bring it out.

04:47 As a constant and then you just have this eye right here and then this is a constant so anything that does not carry any eye has to be a constant and so it has to come out so here it's just going to be one okay because there's no i here so it's just going to be one now remember uh from our previous uh tutorial we said said that summation i running from let me write that in a different color.

05:22 This estimation i from 1 to n is of i is n plus 1 over 2 remember.

05:33 Then we have to n of i squared as n n plus 1 to n plus 1 over 6, you know these things...

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