- #1

mcastillo356

Gold Member

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- TL;DR Summary
- Trying to understand an introduction to the topic of the integral

Hi PF

There goes the quote:

In this section we are going to consider how to find the area of the region ##R## lying under the graph ##y=f(x)## of a nonnegative-valued, continous function ##f##, above the ##x##-axis and between the vertical lines ##x=a## and ##x=b##, where ##a<b##. (See Figure 5.4.) To accomplish this we proceed as follows. Divide the interval ##[a,b]## into ##n## subintervals by using division points:

$$a=x_0<x_1<x_2<x_3<\cdots<x_{n−1}<x_n=b$$

Denote by ##\Delta x_i## the length of the ##i##-th subinterval ##[x_{i−1},x_i]##:

$$\Delta x_i=x_i−x_{i−1},\qquad(i=1,2,3,...,n)$$

Vertically above each subinterval ##[x_{i−1},x_i]## build a rectangle whose base has length ##\Delta x_i## and whose height is ##f(x_i)##. The area of this rectangle is ##f(x_i)\Delta x_i##. Form the sum of these areas:

$$S_n=f(x_1)\Delta x_1+f(x_2)\Delta x_2+f(x_3)\Delta x_3+\cdots+f(x_n)\Delta x_n=\sum_{i=1}^n f(x_i)\Delta x_i$$

The rectangles are shown shaded in Figure 5.5 for a decreasing function ##f##. For an increasing function, the tops of the rectangles would lie above the graph of ##f## rather than below it. Evidently, ##S_n## is an approximation to the area of the region ##R##, and the approximation gets better as ##n## increases, provided we choose the points ##a=x_0<x_1<\cdots<x_n=b## in such a way that the width ##\Delta x_i## of the widest rectangle approaches zero.

Observe in Figure 5.6, for example, that subdividing a subinterval into two smaller subintervals reduces the error in the approximation by reducing that part of the area under the curve that is not contained in the rectangles. It is reasonable, therefore, to calculate the area ##R## by finding the limit of ##S_n## as ##n\rightarrow{\infty}## with the restriction that the largest of the subinterval widths ##\Delta x_i## must approach zero:

$$\displaystyle\text{Area of} R=\underset{\max\Delta x_i\to 0}{\lim_{n\to\infty}} S_n$$

Sometimes, but not always, it is useful to choose the points ##x_i (0\leq i\leq n)## in ##[a,b]## in such a way that the subinterval lengths ##\Delta x_i## are equal. In this case we have

$$\Delta x_i=\Delta x=\displaystyle\frac{b-a}{n},\qquad x_i=a+i\Delta x=a+\displaystyle\frac{i}{n}(b-a)$$

Only one. It's clear that this is a good introduction. States two facts:

(i) the more we subdivide an interval, the better will be the approximation;

(ii) we have to take limits. Here comes the doubt: we must take limit of ##S_n## as ##n\rightarrow{\infty}## with the restriction that the largest of the subinterval widths ##\Delta x_i## must approach zero. It's ok, but, why must we pick the widest subinterval?

No idea; my opinion is that choosing the most wide subinterval and taking limit to zero of it, this is, subdividing the most the largest one, we can forget about doing it to the whole interval... Is it understandable what I mean? I pray for it.

Greetings!

There goes the quote:

**The Basic Area Problem**In this section we are going to consider how to find the area of the region ##R## lying under the graph ##y=f(x)## of a nonnegative-valued, continous function ##f##, above the ##x##-axis and between the vertical lines ##x=a## and ##x=b##, where ##a<b##. (See Figure 5.4.) To accomplish this we proceed as follows. Divide the interval ##[a,b]## into ##n## subintervals by using division points:

$$a=x_0<x_1<x_2<x_3<\cdots<x_{n−1}<x_n=b$$

Denote by ##\Delta x_i## the length of the ##i##-th subinterval ##[x_{i−1},x_i]##:

$$\Delta x_i=x_i−x_{i−1},\qquad(i=1,2,3,...,n)$$

Vertically above each subinterval ##[x_{i−1},x_i]## build a rectangle whose base has length ##\Delta x_i## and whose height is ##f(x_i)##. The area of this rectangle is ##f(x_i)\Delta x_i##. Form the sum of these areas:

$$S_n=f(x_1)\Delta x_1+f(x_2)\Delta x_2+f(x_3)\Delta x_3+\cdots+f(x_n)\Delta x_n=\sum_{i=1}^n f(x_i)\Delta x_i$$

The rectangles are shown shaded in Figure 5.5 for a decreasing function ##f##. For an increasing function, the tops of the rectangles would lie above the graph of ##f## rather than below it. Evidently, ##S_n## is an approximation to the area of the region ##R##, and the approximation gets better as ##n## increases, provided we choose the points ##a=x_0<x_1<\cdots<x_n=b## in such a way that the width ##\Delta x_i## of the widest rectangle approaches zero.

Observe in Figure 5.6, for example, that subdividing a subinterval into two smaller subintervals reduces the error in the approximation by reducing that part of the area under the curve that is not contained in the rectangles. It is reasonable, therefore, to calculate the area ##R## by finding the limit of ##S_n## as ##n\rightarrow{\infty}## with the restriction that the largest of the subinterval widths ##\Delta x_i## must approach zero:

$$\displaystyle\text{Area of} R=\underset{\max\Delta x_i\to 0}{\lim_{n\to\infty}} S_n$$

Sometimes, but not always, it is useful to choose the points ##x_i (0\leq i\leq n)## in ##[a,b]## in such a way that the subinterval lengths ##\Delta x_i## are equal. In this case we have

$$\Delta x_i=\Delta x=\displaystyle\frac{b-a}{n},\qquad x_i=a+i\Delta x=a+\displaystyle\frac{i}{n}(b-a)$$

**Doubt, question**Only one. It's clear that this is a good introduction. States two facts:

(i) the more we subdivide an interval, the better will be the approximation;

(ii) we have to take limits. Here comes the doubt: we must take limit of ##S_n## as ##n\rightarrow{\infty}## with the restriction that the largest of the subinterval widths ##\Delta x_i## must approach zero. It's ok, but, why must we pick the widest subinterval?

**Attempt**No idea; my opinion is that choosing the most wide subinterval and taking limit to zero of it, this is, subdividing the most the largest one, we can forget about doing it to the whole interval... Is it understandable what I mean? I pray for it.

Greetings!