Is this an error?

[Agent Smith]

In the book I'm referring, $\displaystyle -\int_a ^0 (a - u) du$

Shouldn't it be $\displaystyle -\int_a ^ 0 f(a - u) du$?

It's a step in showing $\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^ a \phi (a - x) dx$

I could use some help. Gracias

 

[Dr.Peterson]

In the book I'm referring, $\displaystyle -\int_a ^0 (a - u) du$

Shouldn't it be $\displaystyle -\int_a ^ 0 f(a - u) du$?

It's a step in showing $\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^ a \phi (a - x) dx$

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]

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

 

[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

$\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^a \phi (a - x) dx$

It starts off by saying, "let $x = a - u$"

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

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

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

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

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


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

 

[mario99]

In the book I'm referring, $\displaystyle -\int_a ^0 (a - u) du$

Shouldn't it be $\displaystyle -\int_a ^ 0 f(a - u) du$?

It's a step in showing $\displaystyle \int_0 ^a \phi (x) dx = \int_0 ^ a \phi (a - x) dx$

I could use some help. Gracias

I think that they mean this:

If

$\displaystyle -\int_{a}^{0} (a - u) \ du = \int_{0}^{a} (a - u) \ du = \int_{0}^{a} x \ dx$

Then

$\displaystyle -\int_{a}^{0} f(a - u) \ du = \int_{0}^{a} f(a - u) \ du = \int_{0}^{a} f(x) \ dx$

Or

$\displaystyle -\int_{a}^{0} \phi(a - u) \ du = \int_{0}^{a} \phi(a - u) \ du = \int_{0}^{a} \phi(x) \ dx$


And since $u$ is just a dummy variable, you are allowed to say:

$\displaystyle -\int_{a}^{0} (a - x) \ dx = \int_{0}^{a} (a - x) \ dx = \int_{0}^{a} x \ dx$


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

$\displaystyle \int_0 ^a \phi(x) dx$

We let $x = a - u$

So $\displaystyle \int_0 ^a \phi(a - u) dx$

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

$\displaystyle \int_a ^0 \phi(a - u) dx = -\int_a ^0 \phi (a - u) du$

$\displaystyle - \int_a ^0 \phi (a - u) dx = \int_0 ^a \phi (a - u) du$

$\displaystyle \int_0 ^a \phi (a - u) du = \int_0 ^a \phi (a - x) dx$

So,

$\displaystyle \int_0 ^ a \phi (x) dx = \int_0 ^ a \phi (a - x) dx$

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

 

[mario99]

$\displaystyle \int_a ^0 \phi(a - u) dx = -\int_a ^0 \phi (a - u) du$

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

$\displaystyle \int_{0}^{a} \phi(x) \ dx = -\int_a ^0 \phi (a - u) \ du$

 

[fresh_42]

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

$\displaystyle \int_a ^b \phi (x) dx = \int_a ^ b \phi (y) dy$ because, says my book, "there is no x" in the definite integral, which is $f(b) - f(a)$. $\displaystyle \int_a ^b \phi (x) dx = f(b) - f(a)$. 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
$$\displaystyle \int_a ^b \phi (x) dx = \int_a ^ b \phi (tree) d(tree)$$but people try to avoid using bushes and trees out of convenience.

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

I used $y$ instead of $u$. As I said, it doesn't matter.

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

 

[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 $\phi(x)$ by $f(x).$

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 $dx$ changes, but the integral limits $a,b$ change as well!

 

[Agent Smith]

So $\displaystyle \int x^2 dx = \frac{x^3}{3} + C$

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

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

$\displaystyle \int_0 ^3 (3 - x)^2 dx = \int_0 ^3 (9 - 2x + x^2) dx = ?$

 

[mario99]

So $\displaystyle \int x^2 dx = \frac{x^3}{3} + C$

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

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

$\displaystyle \int_0 ^3 (3 - x)^2 dx = \int_0 ^3 (9 - 2x + x^2) dx = ?$

Usually, you are given this problem:

$\displaystyle \int_{0}^{3}(3 - x)^2 \ dx$

And you change it to:

$\displaystyle \int_{0}^{3}u^2 \ du$

which is the same as:

$\displaystyle \int_{0}^{3}x^2 \ dx$

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

$\displaystyle \int_{0}^{3}(3 - x)^2 \ dx = \int_{0}^{3} 9 - 6x + x^2 \ dx= 9$

 

[fresh_42]

So $\displaystyle \int x^2 dx = \frac{x^3}{3} + C$

Yes.

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

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

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

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

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

$\displaystyle \int_0 ^3 (3 - x)^2 dx = \int_0 ^3 (9 - 2x + x^2) dx = ?$

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

 

[Agent Smith]

@mario99 & @fresh_42

You mean to say $\displaystyle \int_0 ^a \phi (a - x) dx$ "simplifies" to $\displaystyle \int_0 ^a \phi (x) dx$

 

[Agent Smith]

Yes.

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



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

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

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

Gives the same answer, right?

 

[fresh_42]

@mario99 & @fresh_42

You mean to say $\displaystyle \int_0 ^a \phi (a - x) dx$ "simplifies" to $\displaystyle \int_0 ^a \phi (x) dx$

Yes.

 

[fresh_42]

Gives the same answer, right?

Yes, but using $+C$ in definite integrals is ambiguous. It isn't a specific value. It stands for some value in indefinite integrals if we write
$$\int \phi(x)\, dx =f(x)+C.$$The definite version says
$$\int_a^b \phi(x)\, dx =f(b)-f(a).$$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 $+C.$ Let $f(x)$ be the anti-derivative of $\phi(x).$ Then by the fundamental theorem of calculus $$\int_0^x \phi(t)\,dt =f(x)-f(0).$$ We abbreviate it by writing $$\int \phi(x)\,dx=f(x) +C$$ where $C=-f(0) ,$ a value we cannot know unless $f(x)$ is determined. In this sense, $+C$ is "some value". It can be $C=-f(0)$ as above, or $C=-f(1) ,$ another value, if we had considered, e.g. $$\int_1^x \phi(t)\,dt =f(x)-f(1).$$

 

[mario99]

@mario99 & @fresh_42

You mean to say $\displaystyle \int_0 ^a \phi (a - x) dx$ "simplifies" to $\displaystyle \int_0 ^a \phi (x) dx$

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

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