## $f,g \in [0,1] \times [0,1]$, $\int f – g \mathrm{d}x = 0$ and are monotonically increasing, then $\int |f-g| \mathrm{d}x \le \frac{1}{2}$

$$f,g$$ are monotonically increasing in $$[0,1]$$ and $$0\le f , g \le 1$$. $$\int_0^1 f – g \mathrm{d}x = 0$$. Prove that

$$\int_0^1 |f – g|\mathrm{d}x \le \frac{1}{2}$$

In my previous question, $$g(x) = x$$. And my teacher said $$x$$ can be replaced by $$g(x)$$. In fact, in previous question, we don’t need to use the condition $$\int_0^1 f – g \mathrm{d}x = 0$$. But if we replace $$x$$ with $$g$$, this condition becomes necessary.

Also, if $$g = x$$, we can replace the $$\frac{1}{2}$$ with $$\frac{1}{4}$$,that is

$$\int_0^1|f-g| \mathrm{d}x \le \frac{1}{4}$$ I am wondering how to prove that.

## TypeError: ‘int’ object is not subscriptable

How to solve this problem

d1={‘key1’:[1,2,3],(‘key2’):[4,5,6]} d1[‘key2’][0][2]

I want the output as 4,6

## show that $\int \int_{S}^{}{curl\vec{F}\cdot \vec{dS}}$ is proportional to the lenght of $C$

The curve $$C$$ is the edge of a surface S , with $$\vec{T}$$ unit tangent vector of the curve $$C$$ and $$\vec{F}$$ a vector field such as $$\vec{F}=k\vec{T}$$ for each point of $$C$$ , where $$k$$ is a constant , how can I prove that

$$\int \int_{S}^{}{curl\vec{F}\cdot \vec{dS}}\;$$ is proportional to the length of $$C$$

Any help would be appreciate . Thanks in advance.

## If $f$ belongs to $M^{+}$ and $c \ge 0$ then $cf$ belongs to $M^{+}$ and $\int cf = c\int f$

If $$f$$ belongs to $$M^{+}$$ and $$c \ge 0$$ then $$cf$$ belongs to $$M^{+}$$ and $$\int cf = c\int f$$.

I need to proove that, using the following observation:

if $$f\in M^{+}$$ and $$c>0$$, then the mapping $$\varphi \rightarrow \psi = c\varphi$$ is a one-toone mapping between simple function $$\varphi \in M^{+}$$ with $$\varphi \le f$$ and simple functions $$\varphi$$ in $$M^{+}$$ with $$\psi \le cf$$.

I know that this question is already answer here:One-to-one mapping of simple functions $\phi \to \psi = c\,\phi$ implies $\int cf\,d\mu = c \int f\,d\mu$ ?

But I can’t follow the verbal explanation.

My original idea was to proove $$c \int f \le \int cf \le c\int f$$ But I can’t… some idea?

## Integral question $\int x \sqrt {1-x}\ dx.$

So there are (from my knowledge) $$2$$ ways of solving for $$\int x \sqrt {1-x}\ dx.$$

The first is by $$u$$ substitution and the second is by parts. They both differ now i’m confused which one is correct? I have both the correct answers but they differ?

## (local variable) int PlaceNumberValue Error: Cannot Implicity convert type ‘int’ to ‘string’

Buenas noches les dejo este pequeño error no puedo convertir la variable ‘PlaceNumberValue’ de string a entero aqui les dejo el codigo:

using System; using System.Collections.Generic; using System.Linq; using System.Text;

namespace TrackerLibrary { /// /// Represent what the prize if for the given place. /// public class PrizeModel { /// /// The unique identifier for the prize /// public int Id { get; set; } /// /// The numeric identifier for the place(for the second place, etc.) /// public int PlaceNumber { get; set; } /// /// The friendly name for the place (second place, first runner up, etc.) ///

    public string PlaceName { get; set; }      /// <summary>     /// The find amount this place earns or zero if it is not used.     /// </summary>      public decimal PrizeAmount { get; set; }     /// <summary>     /// The number that represent the percentage of theoverall take or     /// zero if it is not used. The percentage if a fraction of 1 (so O.S for     /// 50%)     /// </summary>      public double PrizePercentage { get; set; }      public PrizeModel()     {      }      public PrizeModel(string placeName, string placeNumber, string placeAmount, string placePercentage)     {         PlaceName = PlaceName;         int placeNumberValue = 0;         int.TryParse(placeNumber, out placeNumberValue);         placeNumber = placeNumberValue; //AQUI ESTA EL ERROR          decimal prizeAmountValue = 0;         decimal.TryParse(prizeAmount, out prizeAmountValue);         PrizeAmount = prizeAmountValue;     }       public string prizeAmount { get; set; } } 

}

## Let $(X , \cal{A}, m)$ be a measure space. Let $f:X \to [0,1]$ be measurable. If $m(X) < \infty$, find$\lim_{n\to\infty} \int f^n \, d m$.

Let $$(X , \cal{A}, m)$$ be a measure space. Let $$f:X \to [0,1]$$ be a measurable function. If $$m(X) < \infty$$, determine $$\lim_{n\to\infty} \int f^n \, d m$$.

So far I have:

If $$f(x) < 1$$, then $$\lim_{n\to \infty}{f^n(x)} = 0$$. If $$f(x) = 1$$, then $$\lim_{n\to\infty}{f^n(x)} =1$$. So, for each $$x \in X$$, $$\lim_{n\to\infty}{f^n(x)} = \chi_{_{[f = 1]}}(x).$$

However, I am stuck because I cannot use the Lebesgue Monotone Convergence Theorem, since the sequence is decreasing. Also, I do not know where I will use the hypothesis that $$X$$ is a finite measure space. Any ideas?

## Prove or disprove sentence about $\int f$

I post this question a few months ago. I solved the items (1) and (3), but I cannot to solve (2). Today I read this question again and I’m curious about solution of (2). Today, I had an idea, but I dont know if it works.

Idea. A monotone function has only jump discontinuities. So, $$f|_{[f(x_{0}^{-}),f(x_{0}^{+})]}$$ is continuous, then there is a maximum and minimum. If $$w$$ is a minimum on $$[f(x_{0}^{-}),f(x_{0}^{+})]$$, then

$$w \leq f(x) \Longrightarrow w(x-x_{0}) \leq \int_{x_{0}}^{x}f(t)dt = F(x) – F(x_{0})$$

But, this works for $$f$$ on $$[f(x_{0}^{-}),f(x_{0}^{+})]$$. What about the general case?

## Integration of \int (2x^2 +5) xdx without using integration by parts [on hold]

How to integrate \int (2x^2 +5) xdx without using integration by parts

## TypeError: unsupported operand type(s) for -: ‘int’ and ‘tuple’ – Subtração de dois elementos do vetor [pendente]

Olá, estou tentando subtrair dois elementos de duas listas diferentes, de acordo com as interações mas sempre está mostrando esse erro:

TypeError: unsupported operand type(s) for -: ‘int’ and ‘tuple’

Aqui está meu código:

 for i in range(0, Tam):      dist.append(sqrt(pow(R[i+1] - Rf[i], 2) + pow(G[i+1] - Gf[i], 2)))