Is this an error?

[Agent Smith]

In the book I'm referring, [imath]\displaystyle -\int_a ^0 (a - u) du[/imath]

Shouldn't it be [imath]\displaystyle -\int_a ^ 0 f(a - u) du[/imath]?

It's a step in showing [imath]\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^ a \phi (a - x) dx[/imath]

I could use some help. Gracias

 

[Dr.Peterson]

In the book I'm referring, [imath]\displaystyle -\int_a ^0 (a - u) du[/imath]

Shouldn't it be [imath]\displaystyle -\int_a ^ 0 f(a - u) du[/imath]?

It's a step in showing [imath]\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^ a \phi (a - x) dx[/imath]

I could use some help. Gracias

Context please?

Did you just make up "f", or is it something that seems to have dropped out by a typo? We need to see the whole proof to judge part of it.

 

[fresh_42]

[math]\begin{array}{lll} \int_0^a \phi(a-x)\,dx&=\int_{x=0}^{ x=a}\phi(a-x)\,dx\quad\ ,\ \quad y:=a-x\, , \,\dfrac{dy}{dx}=-1\Longrightarrow dx=-dy\\[6pt] &=-\int_{y=a}^{ y=0}\phi(y)\,dy\\[6pt] &=\int_{y=0}^{ y=a}\phi(y)\,dy\\[6pt] &=\int_{x=0}^{ x=a}\phi(x)\,dx\\[6pt] &=\int_0^a\phi(x)\,dx \end{array}[/math]

 

[Agent Smith]

Context please?

Did you just make up "f", or is it something that seems to have dropped out by a typo? We need to see the whole proof to judge part of it.

It's a step in some kind of "proof" for

[imath]\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^a \phi (a - x) dx[/imath]

It starts off by saying, "let [imath]x = a - u[/imath]"

Then, it says, when x = a, u = 0 and when x = 0, u = a. This so that the rule [imath]\displaystyle \int_a ^b \phi (x) dx = - \int_b ^a \phi (x) dx[/imath] csn be used.

@fresh_42 seems to know exactly what this is about. See his post.

[math]\begin{array}{lll} \int_0^a \phi(a-x)\,dx&=\int_{x=0}^{ x=a}\phi(a-x)\,dx\quad\ ,\ \quad y:=a-x\, , \,\dfrac{dy}{dx}=-1\Longrightarrow dx=-dy\\[6pt] &=-\int_{y=a}^{ y=0}\phi(y)\,dy\\[6pt] &=\int_{y=0}^{ y=a}\phi(y)\,dy\\[6pt] &=\int_{x=0}^{ x=a}\phi(x)\,dx\\[6pt] &=\int_0^a\phi(x)\,dx \end{array}[/math]

There's this rule number 3, in the proof, that you're using, which is

[imath]\displaystyle \int_a ^b \phi (x) dx = \int_a ^ b \phi (y) dy[/imath] because, says my book, "there is no x" in the definite integral, which is [imath]f(b) - f(a)[/imath]. [imath]\displaystyle \int_a ^b \phi (x) dx = f(b) - f(a)[/imath]. I don't get what this means.


There's also [imath]x = a - u \implies dx = -du[/imath]. Where do we use this information? No idea?

 

[mario99]

In the book I'm referring, [imath]\displaystyle -\int_a ^0 (a - u) du[/imath]

Shouldn't it be [imath]\displaystyle -\int_a ^ 0 f(a - u) du[/imath]?

It's a step in showing [imath]\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^ a \phi (a - x) dx[/imath]

I could use some help. Gracias

I think that they mean this:

If

[imath]\displaystyle -\int_{a}^{0} (a - u) \ du = \int_{0}^{a} (a - u) \ du = \int_{0}^{a} x \ dx[/imath]

Then

[imath]\displaystyle -\int_{a}^{0} f(a - u) \ du = \int_{0}^{a} f(a - u) \ du = \int_{0}^{a} f(x) \ dx[/imath]

Or

[imath]\displaystyle -\int_{a}^{0} \phi(a - u) \ du = \int_{0}^{a} \phi(a - u) \ du = \int_{0}^{a} \phi(x) \ dx[/imath]


And since [imath]u[/imath] is just a dummy variable, you are allowed to say:

[imath]\displaystyle -\int_{a}^{0} (a - x) \ dx = \int_{0}^{a} (a - x) \ dx = \int_{0}^{a} x \ dx[/imath]


If they mean the first or the second, it is the same concept. Why are you confused?

 

[Agent Smith]

@mario99

I'll walk you through the proof

[imath]\displaystyle \int_0 ^a \phi(x) dx[/imath]

We let [imath]x = a - u[/imath]

So [imath]\displaystyle \int_0 ^a \phi(a - u) dx[/imath]

[imath]dx = -du[/imath] AND since when x = a, u = 0 and x = 0, u = a, we have to make the following changes

[imath]\displaystyle \int_a ^0 \phi(a - u) dx = -\int_a ^0 \phi (a - u) du[/imath]

[imath]\displaystyle - \int_a ^0 \phi (a - u) dx = \int_0 ^a \phi (a - u) du[/imath]

[imath]\displaystyle \int_0 ^a \phi (a - u) du = \int_0 ^a \phi (a - x) dx[/imath]

So,

[imath]\displaystyle \int_0 ^ a \phi (x) dx = \int_0 ^ a \phi (a - x) dx[/imath]

Can you take a look at the steps and see if there's anything amiss. Explanations would be awesome.

 

[mario99]

[imath]\displaystyle \int_a ^0 \phi(a - u) dx = -\int_a ^0 \phi (a - u) du[/imath]

Everything is correct, but I prefer to write this step as:

[imath]\displaystyle \int_{0}^{a} \phi(x) \ dx = -\int_a ^0 \phi (a - u) \ du[/imath]

 

[fresh_42]

There's this rule number 3, in the proof, that you're using, which is

[imath]\displaystyle \int_a ^b \phi (x) dx = \int_a ^ b \phi (y) dy[/imath] because, says my book, "there is no x" in the definite integral, which is [imath]f(b) - f(a)[/imath]. [imath]\displaystyle \int_a ^b \phi (x) dx = f(b) - f(a)[/imath]. I don't get what this means.

It doesn't mean a lot. It just says that it does not matter how you name the variable. Any notation will do. I could as well have written
[math]\displaystyle \int_a ^b \phi (x) dx = \int_a ^ b \phi (tree) d(tree)[/math]but people try to avoid using bushes and trees out of convenience.

There's also [imath]x = a - u \implies dx = -du[/imath]. Where do we use this information? No idea?

I used [imath] y [/imath] instead of [imath] u [/imath]. As I said, it doesn't matter.

The essential observation was, that if you set, say [imath] u(x)=a-x [/imath] then yes, [imath] dx=-du [/imath] but also [imath] u(0)=a [/imath] and [imath] u(a)=0. [/imath] This means that not only [imath] dx=-du [/imath] has a minus sign, but also the integration path turns its direction, and with [imath] \displaystyle{\int_a^b =-\int_b^a} [/imath] we have
[math] \int_0^a \phi(a-x)\,dx=-\int_a^0 \phi(u)\, du= -\left(-\int_0^a \phi(u)\,du\right) =+\int_0^a\phi(u)\,du=\int_0^a \phi(x)\,dx[/math]Integrals are oriented volumes, i.e. they distinguish between left and right!
[math] \int_a^bf(x)\,dx=F(b)-F(a)=-(F(a)-F(b))=-\int_b^a f(x)\,dx [/math]

 

[Agent Smith]

@fresh_42

Why isn't \(\displaystyle \displaystyle \int_0 ^a \phi (x) = f(a) - f(0)\)

Just like how \(\displaystyle \displaystyle \int_a ^b \phi (x) dx = f(b) - f(a)\)

Also is the following true?

\(\displaystyle \int y (-dx) = - \int y dx\)

 

[fresh_42]

@fresh_42

Why isn't \(\displaystyle \displaystyle \int_0 ^a \phi (x) = f(a) - f(0)\)

It is, if you note the anti-derivative of [imath] \phi(x) [/imath] by [imath] f(x). [/imath]

Just like how \(\displaystyle \displaystyle \int_a ^b \phi (x) dx = f(b) - f(a)\)

Yes.

Also is the following true?

\(\displaystyle \int y (-dx) = - \int y dx\)

Yes, that's true.

However, that wasn't your original question. We had a definite integral and a variable transformation. The crucial point is that if we change the variable, not only the differential [imath] dx [/imath] changes, but the integral limits [imath] a,b [/imath] change as well!

 

[Agent Smith]

So [imath]\displaystyle \int x^2 dx = \frac{x^3}{3} + C[/imath]

I can do this then, [imath]\displaystyle \int_0 ^3 x^2 dx = \left(\frac{3^3}{3} + C\right) - \left(\frac{0^3}{3} + C\right) = 9[/imath]?

I don't have to do this: [imath]\displaystyle \int_0 ^3 x^2 dx = \int_0 ^ 3 (3 - x)^2 dx[/imath]

[imath]\displaystyle \int_0 ^3 (3 - x)^2 dx = \int_0 ^3 (9 - 2x + x^2) dx = ?[/imath]

 

[mario99]

So [imath]\displaystyle \int x^2 dx = \frac{x^3}{3} + C[/imath]

I can do this then, [imath]\displaystyle \int_0 ^3 x^2 dx = \left(\frac{3^3}{3} + C\right) - \left(\frac{0^3}{3} + C\right) = 9[/imath]?

I don't have to do this: [imath]\displaystyle \int_0 ^3 x^2 dx = \int_0 ^ 3 (3 - x)^2 dx[/imath]

[imath]\displaystyle \int_0 ^3 (3 - x)^2 dx = \int_0 ^3 (9 - 2x + x^2) dx = ?[/imath]

Usually, you are given this problem:

[imath]\displaystyle \int_{0}^{3}(3 - x)^2 \ dx[/imath]

And you change it to:

[imath]\displaystyle \int_{0}^{3}u^2 \ du[/imath]

which is the same as:

[imath]\displaystyle \int_{0}^{3}x^2 \ dx[/imath]

And if you do not wanna make a [imath]u[/imath] substitution, you can get rid of the brackets like this:

[imath]\displaystyle \int_{0}^{3}(3 - x)^2 \ dx = \int_{0}^{3} 9 - 6x + x^2 \ dx= 9 [/imath]

 

[fresh_42]

So [imath]\displaystyle \int x^2 dx = \frac{x^3}{3} + C[/imath]

Yes.

I can do this then, [imath]\displaystyle \int_0 ^3 x^2 dx = \left(\frac{3^3}{3} + C\right) - \left(\frac{0^3}{3} + C\right) = 9[/imath]?

No. The [imath] +C [/imath] is only used for indefinite integrals in order to note that integrals are only unique up to additive terms because [imath] (f(x)+C)'=f'(x). [/imath] As soon as we consider definite integrals like
[math] \int_0^3 x^2\,dx =\left[\dfrac{x^3}{3}\right]_0^3 =\dfrac{3^3}{3}-\dfrac{0^3}{3}=9[/math]we don't need [imath] C [/imath] anymore because we have the exact values at both ends of the integral, other than [math] \int x^2 \,dx = \dfrac{x^3}{3}+C[/math] where we only know the anti-derivative, but not the values at its ends. The [imath] +C [/imath] means literally "plus some constant - therefore the [imath] C [/imath] - that we cannot know yet".

I don't have to do this: [imath]\displaystyle \int_0 ^3 x^2 dx = \int_0 ^ 3 (3 - x)^2 dx[/imath]

You can do this, but you don't have to. E.g. if we want to integrate [imath] \int_0^\pi\sin(\pi-x)\,dx [/imath] then we must somehow deal with the [imath] \pi -x [/imath] term. We can do it by using the addition theorem for the sine function, or we can substitute [imath] u=\pi -x [/imath] with [imath] dx=-du [/imath] and [imath] u(0)=\pi\, , \,u(\pi)=0 [/imath] and calculate

[math]\begin{array}{lll} \int_0^\pi\sin(\pi-x)\,dx&=\int_{x=0}^{x=\pi}\sin(\pi-x)\,dx\\[6pt] &=\int_{u=\pi}^{u=0}\sin(u)\,(-du)\\[6pt] &=-\int_{u=\pi}^{u=0}\sin(u)\,du\\[6pt] &=-\left(-\int_{u=0}^{u=\pi}\sin(u)\,du\right)\\[6pt] &=\int_{u=0}^{u=\pi}\sin(u)\,du\\[6pt] &=\left[-\cos(u)\right]_{u=0}^{u=\pi}\\[6pt] &=-\left(\cos(\pi)-\cos(0)\right)\\[6pt] &=-(-1-1)=-(-2)=2 \end{array}[/math]

[imath]\displaystyle \int_0 ^3 (3 - x)^2 dx = \int_0 ^3 (9 - 2x + x^2) dx = ?[/imath]

You can do this by the method we are talking about in this thread and that I just used to directly get [imath]\int_0^3 (3 - x)^2 \,dx=\dfrac{27}{3}=9 [/imath] or do it step by step (with a corrected coefficient)
[math]\begin{array}{lll} \int_0^3 (3 - x)^2 \,dx&= \int_0^3 (9-6x+x^2)\,dx\\[6pt] &=9\int_0^3\,dx-6\int_0^3 x\,dx+\int_0^3x^2\,dx\\[6pt] &=9\left[x\right]_0^3 -6\left[\dfrac{x^2}{2}\right]_0^3+\left[\dfrac{x^3}{3}\right]_0^3\\[16pt] &=9\cdot (3-0) -6 \cdot\left(\dfrac{3^2}{2}-\dfrac{0^2}{2}\right)+\left(\dfrac{27}{3}-\dfrac{0}{3}\right)\\[6pt] &=9\cdot 3- 6\cdot 4.5 + 9\\[6pt] &=9 \end{array}[/math]

 

[Agent Smith]

@mario99 & @fresh_42

You mean to say [imath]\displaystyle \int_0 ^a \phi (a - x) dx[/imath] "simplifies" to [imath]\displaystyle \int_0 ^a \phi (x) dx[/imath]

 

[Agent Smith]

Yes.

No. The [imath] +C [/imath] is only used for indefinite integrals in order to note that integrals are only unique up to additive terms because [imath] (f(x)+C)'=f'(x). [/imath] As soon as we consider definite integrals like
[math] \int_0^3 x^2\,dx =\left[\dfrac{x^3}{3}\right]_0^3 =\dfrac{3^3}{3}-\dfrac{0^3}{3}=9[/math]we don't need [imath] C [/imath] anymore because we have the exact values at both ends of the integral, other than [math] \int x^2 \,dx = \dfrac{x^3}{3}+C[/math] where we only know the anti-derivative, but not the values at its ends. The [imath] +C [/imath] means literally "plus some constant - therefore the [imath] C [/imath] - that we cannot know yet".



You can do this, but you don't have to. E.g. if we want to integrate [imath] \int_0^\pi\sin(\pi-x)\,dx [/imath] then we must somehow deal with the [imath] \pi -x [/imath] term. We can do it by using the addition theorem for the sine function, or we can substitute [imath] u=\pi -x [/imath] with [imath] dx=-du [/imath] and [imath] u(0)=\pi\, , \,u(\pi)=0 [/imath] and calculate

[math]\begin{array}{lll} \int_0^\pi\sin(\pi-x)\,dx&=\int_{x=0}^{x=\pi}\sin(\pi-x)\,dx\\[6pt] &=\int_{u=\pi}^{u=0}\sin(u)\,(-du)\\[6pt] &=-\int_{u=\pi}^{u=0}\sin(u)\,du\\[6pt] &=-\left(-\int_{u=0}^{u=\pi}\sin(u)\,du\right)\\[6pt] &=\int_{u=0}^{u=\pi}\sin(u)\,du\\[6pt] &=\left[-\cos(u)\right]_{u=0}^{u=\pi}\\[6pt] &=-\left(\cos(\pi)-\cos(0)\right)\\[6pt] &=-(-1-1)=-(-2)=2 \end{array}[/math]

You can do this by the method we are talking about in this thread and that I just used to directly get [imath]\int_0^3 (3 - x)^2 \,dx=\dfrac{27}{3}=9 [/imath] or do it step by step (with a corrected coefficient)
[math]\begin{array}{lll} \int_0^3 (3 - x)^2 \,dx&= \int_0^3 (9-6x+x^2)\,dx\\[6pt] &=9\int_0^3\,dx-6\int_0^3 x\,dx+\int_0^3x^2\,dx\\[6pt] &=9\left[x\right]_0^3 -6\left[\dfrac{x^2}{2}\right]_0^3+\left[\dfrac{x^3}{3}\right]_0^3\\[16pt] &=9\cdot (3-0) -6 \cdot\left(\dfrac{3^2}{2}-\dfrac{0^2}{2}\right)+\left(\dfrac{27}{3}-\dfrac{0}{3}\right)\\[6pt] &=9\cdot 3- 6\cdot 4.5 + 9\\[6pt] &=9 \end{array}[/math]

Gives the same answer, right?

 

[fresh_42]

@mario99 & @fresh_42

You mean to say [imath]\displaystyle \int_0 ^a \phi (a - x) dx[/imath] "simplifies" to [imath]\displaystyle \int_0 ^a \phi (x) dx[/imath]

Yes.

 

[fresh_42]

Gives the same answer, right?

Yes, but using [imath] +C [/imath] in definite integrals is ambiguous. It isn't a specific value. It stands for some value in indefinite integrals if we write
[math] \int \phi(x)\, dx =f(x)+C. [/math]The definite version says
[math] \int_a^b \phi(x)\, dx =f(b)-f(a). [/math]There is no place for "some value" in an equation that already has all terms it needs.

 

[fresh_42]

Gives the same answer, right?

There is another way to look at the constant [imath] +C. [/imath] Let [imath] f(x) [/imath] be the anti-derivative of [imath] \phi(x). [/imath] Then by the fundamental theorem of calculus [math] \int_0^x \phi(t)\,dt =f(x)-f(0). [/math] We abbreviate it by writing [math] \int \phi(x)\,dx=f(x) +C [/math] where [imath] C=-f(0) ,[/imath] a value we cannot know unless [imath] f(x) [/imath] is determined. In this sense, [imath] +C [/imath] is "some value". It can be [imath] C=-f(0)[/imath] as above, or [imath] C=-f(1) ,[/imath] another value, if we had considered, e.g. [math] \int_1^x \phi(t)\,dt =f(x)-f(1). [/math]

 

[mario99]

@mario99 & @fresh_42

You mean to say [imath]\displaystyle \int_0 ^a \phi (a - x) dx[/imath] "simplifies" to [imath]\displaystyle \int_0 ^a \phi (x) dx[/imath]

If you will link this to post #11, it is more precisely to say:

[imath]\displaystyle \int_{0}^{a} \phi\left([a - x]^2\right) \ dx = \int_{0}^{a} \phi\left(x^2\right) \ dx[/imath]