Why is this definite integral a single number?

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In summary: Differentiation is accordingly. ##\dfrac{d\,xy}{dy}=x## and we can say "change of area divided by change of width is the height" (pretending the latter is constant as in the...).In summary, the area of the region ##R## is 10 square units.
  • #1
mcastillo356
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Been searching backwards in my textbook, but can't find it; I mean, the reason. I will provide a solved exercise
EXAMPLE 4 Find the area of the region ##R## lying above the line ##y=1## and below the curve ##y=5/(x^2+1)##.

Solution The region ##R## is shaded in Figure 5.24. To find the intersections of ##y=1## and ##y=5/(x^2+1)##, we must solve these equations simultaneously:

##1=\frac{5}{x^2+1}##

so ##x^2+1=5##,##x^2=4##, and ##x=\pm 2##

The area of the region ##R## is the area under the curve ##y=5/(x^2+1)## and above the ##x##-axis between ##x=-2## and ##x=2##, minus the area of a rectangle of width 4 and height 1. Since ##\tan^{-1}x## is an antiderivative of ##1/(x^2+1)##,

##A=\int_{-2}^2 \frac{5}{x^2+1}\ ,dx-4=2\int_0^2\frac{5}{x^2+1}\ ,dx-4=10\tan^{-1} x\left. \right |_0^2-4=10\tan{-1} 2-4## square units.

Observe the use of even symmetry (Theorem 3(h) of Section 5,4) to replace the lower limit of integration by 0. It si easier to susbstitute 0 into the antiderivative than -2.

THEOREM 3 Let ##f## and ##g## be integrable on an interval containing the points ##a##, ##b## and ##c##. Then

(...)

(h) The integral of an even function over an interval symmetric about zero is twice the integral over the positive half of the interval. If ##f## is an even function (i.e., ##f(-x)=f(x))##, then

##\int_{-a}^a f(x)\ ,dx=\int_0^a f(x)\ ,dx## IMG_20230525_104235.jpg

Attempt: It is not explicited the unit
 
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  • #2
What is your question? The one in the title? And if so, why did you add the example?

If ##F(x)## is the anti-derivative of ##f(x)## then ##\displaystyle{\int_a^b\,f(x)\,dx=F(b)-F(a)}.## This is the fundamental theorem of calculus. The LHS is a definite integral and the RHS is a difference between two numbers, hence again a number.

The answer to your question is the fundamental theorem of calculus.
 
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  • #3
This is a consequence of the definition of the Darboux integral.

The upper and lower sums of a real-valued function with respect to a partition are real numbers by closure of addition and multiplication. The lower integral, as the supremum of the lower sums, is a real number by the least upper bound axiom; the upper integral, as the infimum of the upper sums, is a real number by the same axiom. A function is integrable if and only if these two numbers are equal, in which case the value of the integral is that number.
 
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  • #4
mcastillo356 said:
The area of the region ##R## is the area under the curve ##y=5/(x^2+1)## and above the ##x##-axis between ##x=-2## and ##x=2##, minus the area of a rectangle of width 4 and height 1. Since ##\tan^{-1}x## is an antiderivative of ##1/(x^2+1)##,

##A=\int_{-2}^2 \frac{5}{x^2+1}\ ,dx-4=2\int_0^2\frac{5}{x^2+1}\ ,dx-4=10\tan^{-1} x\left. \right |_0^2-4=10\tan{-1} 2-4## square units.

Not the answer to the question you asked, but a comment on your LaTeX. The very last number in your solution should be ##10\tan^{-1}2 - 4##. Omitting the '^' symbol makes what you wrote 10 tan -12 -4.
 
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  • #5
Hi, PF

Mark44 said:
Not the answer to the question you asked, but a comment on your LaTeX. The very last number in your solution should be ##10\tan^{-1}2 - 4##. Omitting the '^' symbol makes what you wrote 10 tan -12 -4.
Perfect.

fresh_42 said:
What is your question? The one in the title?
Got it. I quote "Calculus, 7th ed, R. Adams & C. Essex"

(...) while the definite integral is a pure number, an area is a geometric quantity that impllicitly involves units. If the units along the ##x-## and ##y-##axes are, for example, metres, the area should be quoted in square metres (##m^2##). If units of length along the ##x-##axis and ##y-##axis are not specified, areas should be quoted in square units.

pasmith said:
This is a consequence of the definition of the Darboux integral.

The upper and lower sums of a real-valued function with respect to a partition are real numbers by closure of addition and multiplication. The lower integral, as the supremum of the lower sums, is a real number by the least upper bound axiom; the upper integral, as the infimum of the upper sums, is a real number by the same axiom. A function is integrable if and only if these two numbers are equal, in which case the value of the integral is that number.
Thanks! A point of view of Darboux I haven't noticed untill now. Dense and precise.

Greetings!
 
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  • #6
mcastillo356 said:
Got it. I quote "Calculus, 7th ed, R. Adams & C. Essex"
(...) while the definite integral is a pure number, an area is a geometric quantity that implicitly involves units. If the units along the ##x##− and ##y##−axes are, for example, meters, the area should be quoted in square meters (##m^2##). If units of length along the##x##−axis and ##y##−axis are not specified, areas should be quoted in square units.

You can look at it as $$
\int_a^b y\;dx=[y\cdot x]_{x=a}^{x=b}=(b-a)\cdot y=\underbrace{\int_{from\,[m]}^{to\,[m]}}_{\text{limits of width}} \underbrace{\,height\,[m]}_{=y} \cdot \underbrace{change \;in\;}_{=d} \underbrace{width\,[m]}_{=x} = area\,[m^2]$$
This is the simplest example and everything else are piecewise generalizations such that in the end, we get arbitrary integrals, up to integral curves (flows) through vector fields. It is a long way from width times height to meteorology!

Differentiation is accordingly. ##\dfrac{d\,xy}{dy}=x## and we can say "change of area divided by change of width is the height" (pretending the latter is constant as in the integral).
 
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  • #7
Hi, PF, @fresh_42

fresh_42 said:
$$
\int_a^b y\;dx=[y\cdot x]_{x=a}^{x=b}=(b-a)\cdot y=\underbrace{\int_{from\,[m]}^{to\,[m]}}_{\text{limits of width}} \underbrace{\,height\,[m]}_{=y} \cdot \underbrace{change \;in\;}_{=d} \underbrace{width\,[m]}_{=x} = area\,[m^2]$$

Is it possibly to see it entirely with algebra? Some subjects like dimensional analysis, differentiation, integration, are showed together. Very interesting.

fresh_42 said:
Differentiation is accordingly. ##\dfrac{d\,xy}{dy}=x## and we can say "change of area divided by change of width is the height" (pretending the latter is constant as in the integral).

##\dfrac{d\,xy}{dy}=x## makes me wonder. Why did it takes to get infinitesimals?.

fresh_42 said:
This is the simplest example and everything else are piecewise generalizations such that in the end, we get arbitrary integrals, up to integral curves (flows) through vector fields. It is a long way from width times height to meteorology!

I've browsed Google without success. Vector fields are related with meteorology, but how do I link it all together?
 
  • #8
mcastillo356 said:
Is it possibly to see it entirely with algebra? Some subjects like dimensional analysis, differentiation, integration, are showed together. Very interesting.
It is an analytical concept and the definitions require limits. For a purely algebraic point of view, you would first have to establish such an environment so that the terms make sense. You can consider a differentiation as a derivation (no typo!) or a differential operator, but the corresponding integral operator uses an integral in its definition, i.e. again an analytical term.

mcastillo356 said:
##\dfrac{d\,xy}{dy}=x## makes me wonder. Why did it takes to get infinitesimals?
The infinitesimal notation - in this context - is more of a reminder not to forget that we deal with limits (change of) rather than having a meaning of their own.
$$\left. \dfrac{d}{d\,x}\right|_{x_0}f(x)=\left. \dfrac{d\,f}{d\,x}\right|_{x_0} =\displaystyle{\lim_{h \to 0}}\dfrac{f(x_0+h)-f(x_0)}{h}.$$
You cannot discuss them separately. Best you can get is Weierstrass's formula
$$
f(x_0+h)=f(x_0)+ \underbrace{\left(\left. \dfrac{d}{d\,x}\right|_{x_0}f(x)\right)}_{=D_{x_0}f}\cdot h + r(h)
$$
where all the infinitesimal quantities, all limits, anything going to zero, all those are put into the correction term ##r(h)## which is required to converge faster to zero than ##h## does, and ##D_{x_0}f## is merely a linear function and not the quotient of infinitesimals.

In other contexts, especially in physics, ##dx## can also mean a basis vector of the cotangent space or a differential, a Pfaffian form. These cases do not use quotients of ##d## expressions and can thus be explained without limits, or better: with hidden limits since we still use tangents.
mcastillo356 said:
I've browsed Google without success. Vector fields are related with meteorology, but how do I link it all together?
That was more of a rhetorical statement. Meteorology is full of (tangent) vector fields. You see at least one whenever you see a weather forecast, often literally with drawn vectors. Yet, the way from a bunch of secants converging to a tangent up to the theorem of the hedgehog as we call the hairy ball theorem is long, or to meteorology.

Here (pretty much at the beginning) is a list of 10 perspectives on the differentiation process:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
and I didn't even use the word slope.

You have to go this way if you study physics, at least the first steps to cotangent spaces and differential forms. If you want to study physics on a deeper level, then you will probably need pushforwards and pullbacks, too.
 
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  • #9
Hi, PF, @fresh_42, I think I got it:

fresh_42 said:
You can look at it as $$
\int_a^b y\;dx=[y\cdot x]_{x=a}^{x=b}=(b-a)\cdot y=\underbrace{\int_{from\,[m]}^{to\,[m]}}_{\text{limits of width}} \underbrace{\,height\,[m]}_{=y} \cdot \underbrace{change \;in\;}_{=d} \underbrace{width\,[m]}_{=x} = area\,[m^2]$$
It is as simple as I didn't think untill now. The only obstacle is the fact that in dimensional analisis both height and width have the same expression: ##[L]##: Area=##[L]^2##.
I truly thank the quote, that so precisely makes the path between the definite integral (a pure number), and the area, a dimensional quantity
fresh_42 said:
Differentiation is accordingly. ##\dfrac{d\,xy}{dy}=x## and we can say "change of area divided by change of width is the height" (pretending the latter is constant as in the integral).
Just perfect.

Unfortunately, I cannot discuss any of the other grounds mentioned in the thread, but I can say that my doubt is solved.

Thanks!
 
  • #10
Just thanks to Fresh about the answers to what seems a simple question but is surprisingly subtle. A study of real analysis would help a lot. It becomes more intuitively obvious using infinitesimals. I am working on a surprising update to my insights article on infinitesimals and a supplement to an Algebra 2 Text that gives a calculus introduction using infinitesimals.

To the OP these days the rigorous treatment of calculus using infinitesimals is known to be logically equivalent to calculus using limits. The choice of which you use is just personal taste.

Thanks
Bill
 
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  • #11
Strange a bit that a direct answer has not been appeared in the thread.

Why is the definite integral a single number?​


By definition
 
  • #12
You can always take it to be a function of the endpoints:$$I(a,b) = \int_a^b f(x) \ dx$$Where $$I: \mathbb R \times \mathbb R \rightarrow \mathbb R$$
 
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  • #13
PeroK said:
You can always take it to be a function of the endpoints
Yes, and a linear functional on the space of Riemann integrable functions, etc.
But I believe that it would be nice to mention the definition first. Riemann sums, tagged partitions you know.
 
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  • #14
wrobel said:
Yes, and a linear functional on the space of Riemann integrable functions, etc.
But I believe that it would be nice to mention the definition first. Riemann sums, tagged partitions you know.

See post #3.
 
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  • #15
PeroK said:
You can always take it to be a function of the endpoints:$$I(a,b) = \int_a^b f(x) \ dx$$Where $$I: \mathbb R \times \mathbb R \rightarrow \mathbb R$$
Sorry for the delay, the truly reason is that I didn't understand the mapping of the cartesian product into the real numbers.
 
  • #16
pasmith said:
This is a consequence of the definition of the Darboux integral.

The upper and lower sums of a real-valued function with respect to a partition are real numbers by closure of addition and multiplication. The lower integral, as the supremum of the lower sums, is a real number by the least upper bound axiom; the upper integral, as the infimum of the upper sums, is a real number by the same axiom. A function is integrable if and only if these two numbers are equal, in which case the value of the integral is that number.
Darboux is equivalent to Riemann, the path I'm following. But easier.
 
  • #17
mcastillo356 said:
Sorry for the delay, the truly reason is that I didn't understand the mapping of the cartesian product into the real numbers.
Differentiation and therefore integration, too, has many different perspectives, depending on what you regard as variables. In the case of integration, we have the function ##f(x)##, the integration variable ##dx##, and the integration limits ##a## and ##b##. E.g.:

##f\longmapsto \int f(x)\,dx## is a functional or an integral operator, the machine that integrates.

##x\longmapsto \int_0^x f(t)\,dt ## is a function in one variable ##x##, the area under the graph between ##0## and ##x##.

##(a,b)\longmapsto \int_a^b f(x)\,dx## is a function in two variables, the machine that computes those (oriented) areas.

In higher dimensions, we also get directions.
 
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  • #18
mcastillo356 said:
Sorry for the delay, the truly reason is that I didn't understand the mapping of the cartesian product into the real numbers.
Take any two real numbers ##a, b## and compute the integral ##I(a, b)## and you get a real number. That's a mapping, no? Note that in this case we take ##f## to be a fixed function.
 
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  • #19
mcastillo356 said:
Sorry for the delay, the truly reason is that I didn't understand the mapping of the cartesian product into the real numbers.
Note that it's more usual to take ##[a, b]## to be a fixed interval and, as @wrobel says, view the integral as a linear functional (which is a specific type of mapping) from the set of integrable functions to the real numbers. In this case, we have:
$$I: \mathbb L \rightarrow \mathbb R$$Where$$I(f) = \int_a^b f(x) \ dx$$This idea of using the definite integral as a linear functional has far-reaching applications in mathematical physics.
 
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  • #20
PeroK said:
, view the integral as a linear functional (which is a specific type of mapping) from the set of integrable functions to the real numbers. In this case, we have:
$$I: \mathbb L \rightarrow \mathbb R$$Where$$I(f) = \int_a^b f(x) \ dx$$This idea of using the definite integral as a linear functional has far-reaching applications in mathematical physics.
Fine. A quick look at Wikipedia has give me a good introduction. But I've noticed a background lack on algebraic structures.
Greetings
 
  • #21
mcastillo356 said:
Fine. A quick look at Wikipedia has give me a good introduction. But I've noticed a background lack on algebraic structures.
Greetings
The question "What is an integral?" can guide you through the entire curriculum of analysis. It reaches from simple calculations of areas and rotational volumes to complex analysis, measure theory, and differential geometry.
 

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