Odd Integration Proof (1 Viewer)

ProdigyInspired · Aug 6, 2016

So, as most Ext 2 students know, the last part of Integration involves proving the odd and even definite function proofs.
So for the odd proof (<s>I'll just do it below </s>, its completely wrong don't follow)
<s>
$Show\quad \int _{ -a }^{ a }{ f(x) } dx\quad =\quad 0$

$Let\quad u\quad =\quad -x\\ \\ \begin{matrix} When\quad x\quad =\quad a,\quad u\quad =\quad -a \\ when\quad x\quad =\quad -a,\quad u\quad =\quad a \end{matrix}\\ \begin{matrix} \therefore \quad I\quad =\quad \int _{ -a }^{ a }{ f(-u) } (-du) \\ =\quad -\int _{ -a }^{ a }{ f(u) } (du) \end{matrix}$

Then I get stuck here.

$Switch\quad variables\quad of\quad u\quad with\quad x\\ -\int _{ 0 }^{ a }{ f(u) } =-\int _{ 0 }^{ a }{ f(x) }$ </s>

I've done it before but I still can't wrap my head around the concept of the dummy variable of x.

If the area is bound by a curve that uses the value of x i.e. f(x), why would it be the same thing as using a variable of f(u)?

ProdigyInspired · Aug 6, 2016

I've realised I messed up my working out, but I guess that's not that important. I can get to the splitting part, and I think a better way is to convert one of them into a negative value which makes it = 0.

InteGrand · Aug 6, 2016

ProdigyInspired said:
So, as most Ext 2 students know, the last part of Integration involves proving the odd and even definite function proofs.
So for the odd proof (I'll just do it below)

$Show\quad \int _{ -a }^{ a }{ f(x) } dx\quad =\quad 0$

$Let\quad u\quad =\quad -x\\ \\ \begin{matrix} When\quad x\quad =\quad a,\quad u\quad =\quad -a \\ when\quad x\quad =\quad -a,\quad u\quad =\quad a \end{matrix}\\ \begin{matrix} \therefore \quad I\quad =\quad \int _{ -a }^{ a }{ f(-u) } (-du) \\ =\quad -\int _{ -a }^{ a }{ f(u) } (du) \end{matrix}$

Then I get stuck here.

$Switch\quad variables\quad of\quad u\quad with\quad x\\ -\int _{ 0 }^{ a }{ f(u) } =-\int _{ 0 }^{ a }{ f(x) }$

I've done it before but I still can't wrap my head around the concept of the dummy variable of x.

If the area is bound by a curve that uses the value of x i.e. f(x), why would it be the same thing as using a variable of f(u)?

One way to think of it is that integral from x = 0 to x = a of f(x) dx is the (signed) area under the graph of y = f(x).

The integral from u = 0 to u = a of f(u) du is the signed area under the graph of y = f(u).

These two graphs are the same, since they have the same equation and bounds; the graph of say y = sin(x) between x = 0 and pi looks exactly the same as the graph of y = sin(u) between u = 0 and u = pi. In other words, the name given to the independent variable of course doesn't change the graph shape, the only thing affecting this is the function itself, which is exactly the same in both cases, with same bounds.

Since the graphs are identical and the bounds are identical, the areas in question (and hence the integrals) are identical.

ProdigyInspired · Aug 6, 2016

InteGrand said:
One way to think of it is that integral from x = 0 to x = a of f(x) dx is the (signed) area under the graph of y = f(x).

The integral from u = 0 to u = a of f(u) du is the signed area under the graph of y = f(u).

These two graphs are the same, since they have the same equation and bounds; the graph of say y = sin(x) between x = 0 and pi looks exactly the same as the graph of y = sin(u) between u = 0 and u = pi. In other words, the name given to the independent variable of course doesn't change the graph shape, the only thing affecting this is the function itself, which is exactly the same in both cases, with same bounds.

Since the graphs are identical and the bounds are identical, the areas in question (and hence the integrals) are identical.

Does this work the same way with the identity of def int f(x) = def int f(a-x)?

The a-x and x are essentially the same, as they give the area stays exactly the same.

But how does that work when say, changing the a variable to a value that makes the area, inspected visually, far smaller than the original graph?

InteGrand · Aug 6, 2016

ProdigyInspired said:
Does this work the same way with the identity of def int f(x) = def int f(a-x)?

The a-x and x are essentially the same, as they give the area stays exactly the same.

Yes that dummy variables idea is used there too.

$\noindenf In general, $\int _a ^b f(t)\, \mathrm{d}t$ means \emph{exactly the same thing} as $\int _a ^b f(x) \, \mathrm{d}x$. It's the actual function and bounds that matter, not the symbol used inside the function. For this reason, we could just write the integrals as $\int _a ^b f$ (but then we'd need to separately write the definition of $f$ every time, which'd get tedious sometimes.)$

You may also want to see this thread and the post by Carrotsticks: http://community.boredofstudies.org...ion-2/342339/dummy-variables.html#post7030860 .

ProdigyInspired · Aug 6, 2016

InteGrand said:
Yes that dummy variables idea is used there too.

$\noindenf In general, $\int _a ^b f(t)\, \mathrm{d}t$ means \emph{exactly the same thing} as $\int _a ^b f(x) \, \mathrm{d}x$. It's the actual function and bounds that matter, not the symbol used inside the function. For this reason, we could just write the integrals as $\int _a ^b f$ (but then we'd need to separately write the definition of $f$ every time, which'd get tedious sometimes.)$

You may also want to see this thread and the post by Carrotsticks: http://community.boredofstudies.org...ion-2/342339/dummy-variables.html#post7030860 .

I see.

I understand that any variable can be the same, provided the boundaries are also met.

In the respect of in particularly A( f(a-x) ) = A( f(x) ), how does that work then? I understand the shape of the function never changes unless a different function is used, however the Area does fluctuate. Using y = x^3 as an example, the area from 0 to 2 in y = x^3, does not equal the area between to 0 to 2 in y = 2 - x^3.

Does this mean that the boundaries differ based on the values of a? Or do they remain the same?

ProdigyInspired · Aug 6, 2016

<s> Would a viable explanation be that, with every change in the x, i.e. the movement caused by the a and negative, the boundaries are changed? However despite their change they are still essentially the same boundaries. Similarly to how the area between 0 to 2 in x^3 would be the same as 4 to 2 in 2 - x^3.

But if the integrals are listed with the same boundaries, how does this work? </s>

InteGrand · Aug 6, 2016

ProdigyInspired said:
I see.

In the respect of in particularly A( f(a-x) ) = A( f(x) ), how does that work then? I understand the shape of the function never changes unless a different function is used, however the Area does fluctuate. Using y = x^3 as an example, the area from 0 to 2 in y = x^3, does not equal the area between to 0 to 2 in y = 2 - x^3.

Does this mean that the boundaries differ based on the values of a? Or do they remain the same?

$$\noindent Well first of all, in that example, $f(2-x)$ isn't $2-x^3$, it's $\left(2-x\right)^3$. This indeed has the same area from $0$ to $2$.$

$\noindent Let $I = \int _0 ^a f(x)\, \mathrm{d}x$. Let $u=a-x$, then we get $I = \int _a ^ 0 f(a-u) \, \cdot (-1) \, \mathrm{d}u = \int _0 ^a f(a-u)\, \mathrm{d}u$. Again by dummy variables, this is the same thing as $\int _0 ^ a f(a-x)\, \mathrm{d}x$, so this is also $I$, and the proof is complete. In other words, if we use dummy variables here, it's to talk about $f(a-u)$ and $f(a-x)$.$

InteGrand · Aug 6, 2016

ProdigyInspired said:
Would a viable explanation be that, with every change in the x, i.e. the movement caused by the a and negative, the boundaries are changed? However despite their change they are still essentially the same boundaries. Similarly to how the area between 0 to 2 in x^3 would be the same as 4 to 2 in 2 - x^3.

But if the integrals are listed with the same boundaries, how does this work?

$\noindent Well as said above (which I know was posted later than your comment), $2-x^3$ is irrelevant to the discussion. What $2-x^3$ is is $2-f(x)$, rather than $f(2-x)$.$

$\noindent Whenever we use dummy variables in these integrals, we're just saying that if the function is the \emph{same} and the bounds are the same, the variable of integration used is essentially irrelevant to the value of the integral. So dummy variables aren't used to compare the integral of $f(x)$ vs. $f(a-x)$ (which are two \emph{different} functions), it's just used by some of the textbooks to `justify' why they just changed $u$'s to $x$'s in an integral and kept everything else the same.$

ProdigyInspired · Aug 6, 2016

InteGrand said:
$$\noindent Well first of all, in that example, $f(2-x)$ isn't $2-x^3$, it's $\left(2-x\right)^3$. This indeed has the same area from $0$ to $2$.$

$\noindent Let $I = \int _0 ^a f(x)\, \mathrm{d}x$. Let $u=a-x$, then we get $I = \int _a ^ 0 f(a-u) \, \cdot (-1) \, \mathrm{d}u = \int _0 ^a f(a-u)\, \mathrm{d}u$. Again by dummy variables, this is the same thing as $\int _0 ^ a f(a-x)\, \mathrm{d}x$, so this is also $I$, and the proof is complete. In other words, if we use dummy variables here, it's to talk about $f(a-u)$ and $f(a-x)$.$

Ah my bad, a better example of what I was thinking would be f(3-x). i.e. y = (3-x)^3

So are visual sketches completely thrown out the window when considering such areas? I now understand the algebraic proof, but my doubts still preside when the two graphs are seen visually, and their areas.

InteGrand · Aug 6, 2016

ProdigyInspired said:
Ah my bad, a better example of what I was thinking would be f(3-x). i.e. y = (3-x)^3

So are visual sketches completely thrown out the window when considering such areas? I now understand the algebraic proof, but my doubts still preside when the two graphs are seen visually, and their areas.

$\noindent Dummy variables would only get used by TB's to say something like $\int _0 ^3 \left(3-u\right)^3\, \mathrm{d}u =\int _0 ^3 \left(3-x\right)^3\, \mathrm{d}x$. (In other words, the functions are the \emph{same} and we \emph{only} changed a letter.) Dummy variables wouldn't justify something like $\int _0 ^3 \left(3-u\right)^3 \, \mathrm{d}u = \int _0 ^3 x^3 \, \mathrm{d}x$, since the functions are different.$

Shadowdude · Aug 6, 2016

i think it might be better that instead of saying "switch variables u with x"

to say "let u = x" and pretend it's a substitution

you'll probably find yourself unconfused by the end of that exercise

ProdigyInspired · Aug 7, 2016

Shadowdude said:
i think it might be better that instead of saying "switch variables u with x"

to say "let u = x" and pretend it's a substitution

you'll probably find yourself unconfused by the end of that exercise

that would probably make myself more confused since I had already subbed u = -x

Shadowdude · Aug 7, 2016

ProdigyInspired said:
that would probably make myself more confused since I had already subbed u = -x

Sub in v = x then

Odd Integration Proof (1 Viewer)

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