The Structural analysis

[mishin05]

The classical analysis has errors. That to eliminate them I have thought up the structural analysis.

The basic formula:$$\displaystyle \int f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt$$.

It leads to following contradictions for example:

the Structural analysis:

$$\displaystyle\int(a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}\not=\int(a+x)d(a+x)$$;

$$\displaystyle\int\limits_{0}^{x}2tdt=\int2xdx=x^2$$;

$$\displaystyle\int\limits_{0}^{\sqrt{x^{2}+C}}2tdt=x^2+C$$.

P.S.

1. $$\displaystyle \int\limits_{a}^{a+x}tdt=\frac{(a+x)^2}{2}-\frac{a^2}{2}$$;

2. $$\displaystyle \int (a+x)dx=(a+x)x-\int xd(a+x)$$;

3. $$\displaystyle \int (a+x)dx=(a+x)x-\int xda-\int xdx$$;

4. $$\displaystyle \int (a+x)dx=(a+x)x-\int xdx$$;

5. $$\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\frac{x^2}{2}\right)-\int xdx$$;

6. $$\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}+\int xdx\right)-\int xdx$$;

7. $$\displaystyle \int (a+x)dx=\left(\frac{(a+x)^2}{2}-\frac{a^2}{2}\right)+\left(\int xdx-\int xdx\right)$$;

8. $$\displaystyle \int (a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}$$;

1. and 8. $$\rightarrow$$$$\displaystyle \int (a+x)dx=\int\limits_{a}^{a+x} f(t)dt$$.

 

[Tach]

The classical analysis has errors. That to eliminate them I have thought up the structural analysis.

The basic formula:$$\displaystyle \int f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt$$.

Your starting point is false. The RHS and the LHS are DIFFERENT types of integrals. This leads to the whole slew of errors in your post.

 

[mishin05]

Your starting point is false. The RHS and the LHS are DIFFERENT types of integrals. This leads to the whole slew of errors in your post.

That you have except nothing eaning words. Prove that you confirm!

I have resulted the proof that they are equal. You it is simple so say that they aren't equal. Then you offer not mathematics, and children's dispute!

 

[AlphaNumeric]

I have resulted the proof that they are equal. You it is simple so say that they aren't equal. Then you offer not mathematics, and children's dispute!

They are not in general equal because the left hand side is an indefinite integral, having no limits, while the right hand side has limits. There's other problems too.

$$\int f(a+x)dx$$ is a function of a, not x. The x 'variable' is a dummy variable, it is integrated out. As a result you can replace it with any variable, $$\int f(a+x)dx = \int f(a+y)dy = \int f(a+Jeff) d(Jeff)$$. The right hand side is a function of both a and x, as it doesn't integrate out x, it integrates out the t variable. Thus even if you put limits in the equality would only be possible for particular values of x in the right hand side, as the left hand side is independent of it.

Let's do some actual mathematics to see what your right hand side could be, starting with $$I = \int_{a}^{a+x}f(t) dt$$. Change variables t = y+a, do dt = dy and the limits change as a+x -> a and a -> 0. Thus $$I = \int_{0}^{x}f(y+a)dy$$. Relabelling t to t we find $$\int_{a}^{x+a}f(t)dt = \int_{0}^{x}f(t+a)dt$$

This is the formula you should have started with. You should go back to your high school textbook and learn how to apply limits properly.

 

[Guest254]

Hurrah, you've returned!

This leads to the whole slew of errors in your post.

And on this note... don't you have some apologies you want to start dishing out?

 

[mishin05]

They are not in general equal because the left hand side is an indefinite integral, having no limits, while the right hand side has limits. There's other problems too.

$$\int f(a+x)dx$$ is a function of a, not x. The x 'variable' is a dummy variable, it is integrated out. As a result you can replace it with any variable, $$\int f(a+x)dx = \int f(a+y)dy = \int f(a+Jeff) d(Jeff)$$. The right hand side is a function of both a and x, as it doesn't integrate out x, it integrates out the t variable. Thus even if you put limits in the equality would only be possible for particular values of x in the right hand side, as the left hand side is independent of it.

Let's do some actual mathematics to see what your right hand side could be, starting with $$I = \int_{a}^{a+x}f(t) dt$$. Change variables t = y+a, do dt = dy and the limits change as a+x -> a and a -> 0. Thus $$I = \int_{0}^{x}f(y+a)dy$$. Relabelling t to t we find $$\int_{a}^{x+a}f(t)dt = \int_{0}^{x}f(t+a)dt$$

This is the formula you should have started with. You should go back to your high school textbook and learn how to apply limits properly.

You speak silly. I assert that that is written in high school textbook incorrectly. At uncertain integral limits are put down because they aren't present, that is why that they are defined by an integration variable:$$\int f(x)dx=\int\limits_{0}^{x} f(t)dt$$.
In $$\int f(a+x)dx$$ $$a=const.$$, $$x - variable$$.

Let's admit:
$$a=t_0, x=\Delta t$$, then:

$$\int f(a+x)dx=\int f(t_0+\Delta t)d\Delta t=\int f(t_0+\Delta t)d\left((t_0+\Delta t)-f(t_0)\right)=\int f(t_0+\Delta t)d(t_0+\Delta t)-\int f(t_0+\Delta t)df(t_0)=F(t_0+\Delta t)-F(t_0)$$.

$$\int\limits_{t_o}^{t_0+\Delta t}f(t)dt=F(t_0+\Delta t)-F(t_0)$$

 

[Tach]

Hurrah, you've returned!

And on this note... don't you have some apologies you want to start dishing out?

No,go stalking elsewhere.

 

[Tach]

You speak silly. I assert that that is written in high school textbook incorrectly. At uncertain integral limits are put down because they aren't present, that is why that they are defined by an integration variable:$$\int f(x)dx=\int\limits_{0}^{x} f(t)dt$$.
In $$\int f(a+x)dx$$ $$a=const.$$, $$x - variable$$.

Let's admit:
$$a=t_0, x=\Delta t$$, then:

$$\int f(a+x)dx=\int f(t_0+\Delta t)d\Delta t=\int f(t_0+\Delta t)d\left((t_0+\Delta t)-f(t_0)\right)=\int f(t_0+\Delta t)d(t_0+\Delta t)-\int f(t_0+\Delta t)df(t_0)=F(t_0+\Delta t)-F(t_0)$$.

$$\int\limits_{t_o}^{t_0+\Delta t}f(t)dt=F(t_0+\Delta t)-F(t_0)$$

The LHS is an indefinite integral, the RHS is a definite integral, this is why your "proof" leads to nonsense.

 

[Guest254]

No,go stalking elsewhere.

I'll bump the thread for your convenience.

 

[Tach]

The classical analysis has errors. That to eliminate them I have thought up the structural analysis.

The basic formula:$$\displaystyle \int f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt$$.

Try writing it correctly for a change:

$$\displaystyle \int\limits_{0}^{x} f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt$$

 

[camilus]

The x 'variable' is a dummy variable, it is integrated out. As a result you can replace it with any variable, $$\int f(a+x)dx = \int f(a+y)dy = \int f(a+Jeff) d(Jeff)$$.

hahaha that was awsome, it reminded me of a story I read that Hilbert once famously declared that the theory of geometry would still make sense if points, lines and planes were replaced by tables, chairs and beer mugs.

 

[D H]

Assuming of course you don't drink so many planes that you fall off the line and end up under the point. Then it all becomes pointless.

 

[AlphaNumeric]

$$\int f(x)dx=\int\limits_{0}^{x} f(t)dt$$.

That isn't an identity, it isn't true in general.

In $$\int f(a+x)dx$$ $$a=const.$$, $$x - variable$$.

The expression depends on a, it is not dependent on x. You need to learn basic calculus again. Actually 'again' implies you learnt it before, which you obviously haven't.

 

[mishin05]

Try writing it correctly for a change:

$$\displaystyle \int\limits_{0}^{x} f(a+x)dx=\int\limits_{a}^{a+x} f(t)dt$$

All is true! $$\displaystyle \int\limits_{0}^{x} f(a+x)dx=\int f(a+x)dx$$

$$\displaystyle \int\limits_{0}^{x} f(a+x)dx=ax+\frac{x^2}{2}$$

$$\displaystyle \int f(a+x)dx=ax+\frac{x^2}{2}$$

$$\displaystyle \int f(a+x)dx\not=ax+\frac{x^2}{2}+C$$

$$\displaystyle \int\limits_{a}^{\sqrt{(a+x)+2C}} ftdt=ax+\frac{x^2}{2}+C$$

Any integral has limits. Because differentiation process is made not with function, and with a function increment. Therefore return process - integration leads to an increment (Newton-Lejbnitsa formula) to function. A difference between uncertain integral and defined only that the uncertain integral is limited by an integration variable, and defined - values of this variable!

 

[mishin05]

That isn't an identity, it isn't true in general.

The expression depends on a, it is not dependent on x. You need to learn basic calculus again. Actually 'again' implies you learnt it before, which you obviously haven't.

The best for understanding is a graphic representation. I am ready to show and prove you all on the function schedule. But I don't see here possibility to show the schedule. If know how to place on this site a graphic representation, prompt, please!

 

[D H]

mishin05, please learn the difference between an indefinite and definite integral. All of your problems stem from your own misunderstanding.

 

[prometheus]

As a result you can replace it with any variable, $$\int f(a+x)dx = \int f(a+y)dy = \int f(a+Jeff) d(Jeff)$$.

What's this integral: $$\int \frac{1}{\mathrm{cabin}}d \mathrm{cabin}$$

 

[mishin05]

mishin05, please learn the difference between an indefinite and definite integral. All of your problems stem from your own misunderstanding.

You want, that I have studied that someone has told about the certain and uncertain integral. Because you have got used to repeat another's thoughts.

There is a formula: $$U\cdot V=\int UdV+\int VdU$$!!!

From which it is possible to define very simply that such uncertain integral. It is necessary to have only a little bit mind and independence in reasonings. In the most simple cases in geometrical interpretation the sum of two uncertain integrals is limited by the rectangle area $$U\cdot V$$.
Understand, it is limited! Think independently, instead of repeat delirium of madwomen which write textbooks. They too can be mistaken!

 

[Pete]

What's this integral: $$\int \frac{1}{\mathrm{cabin}}d \mathrm{cabin}$$

 

[D H]

Wrong picture! That cabin should not be on land.

This is much closer to the correct answer: