# Thread: Integration Transformations Proof

1. ## Integration Transformations Proof

I was told this was a rather easy question, but I'm having troubles sorting it all out....

Given that $f$ is an integrable function on $[a,b]$, prove

1) For any $c>0$, $\int_{a}^{b}f(x)dx=c\int_{a/c}^{b/c}f(cx)dx$
2) For any $c\in\mathbb{R}$, $\int_{a}^{b}f(x)dx=\int_{a-c}^{b-c}f(x+c)dx$

The proof is said to be simplest by starting with the direct definitions of integrability.

2. Originally Posted by DonnMega
I was told this was a rather easy question, but I'm having troubles sorting it all out....

Given that $f$ is an integrable function on $[a,b]$, prove

1) For any $c>0$, $\int_{a}^{b}f(x)dx=c\int_{a/c}^{b/c}f(cx)dx$
2) For any $c\in\mathbb{R}$, $\int_{a}^{b}f(x)dx=\int_{a-c}^{b-c}f(x+c)dx$

The proof is said to be simplest by starting with the direct definitions of integrability.
What, particularly, are you having trouble with?

3. This is what I've got so far...

For 1)
taking the definition of an integral, we know that the integral is equal to the supremum of the lower sums and the infimum of the upper sums:
$\int_{a}^{b}f(x)dx=$sup $_P(L(f,P))=$inf $_P(U(f,P))$

taking this a step farther, we can put this into summation notation:
$\int_{a}^{b}f(x)dx=$sup $_P(\sum_{1}^{J}m_j(x_j-x_{j-1}))$

Then it's as easy as seeing that since $f(x)$ has become $f(cx)$, we now have
sup $_P(\sum_{1}^{J}m_j(cx_j-cx_{j-1}))$

then
sup $_P(\sum_{1}^{J}m_jc(x_j-x_{j-1}))$

then
sup $_P(c\sum_{1}^{J}m_j(x_j-x_{j-1}))$

then
$c$ sup $_P(\sum_{1}^{J}m_j(x_j-x_{j-1}))=c\int_{a}^{b}f(cx)dx$

so it's obvious where the $c$ comes from, it's the $a/c$ and $b/c$ of the integral that i can't quite explain

4. So I've finished the proof! YAY! But...I've come up with a slightly different answer. Have i gone wrong somewhere? Is the question wrong? Any comments will help, thank you!

pf//
taking the definition of an integral, we know that the integral is equal to the supremum of the lower sums and the infimum of the upper sums:
$\int f(x)dx=$sup $_P(L(f,P))=$inf $_P(U(f,P))$

taking this a step farther, we can put this into summation notation:
$\int f(x)dx=$sup $_P(\sum_{1}^{J}m_j(x_j-x_{j-1}))$

Now we see that since $f(x)$ becomes $f(cx)$, we essentially have a mapping $x->cx$ and our image is
$\int f(cx)dx=$ sup $_P(\sum_{1}^{J}m_j(cx_j-cx_{j-1}))$

from here we simplify and bring our constant $c$ completely out
sup $_P(\sum_{1}^{J}m_jc(x_j-x_{j-1}))$

sup $_P(c\sum_{1}^{J}m_j(x_j-x_{j-1}))$

$c$ sup $_P(\sum_{1}^{J}m_j(x_j-x_{j-1}))=c\int f(x)dx$

Equating this to the left side of the equation after mapping, we see that
$\int f(cx)dx=c\int f(x)dx$

$\frac {1}{c} \int f(cx)dx=\int f(x)dx$

As for the change in the limits of integration, we first note that our mapping of $x -> cx$ has shifted our interval
$[a,b] -> [ca, cb]$

to obtain our origional interval, we must compose our limits of integration with the inverse mapping function $cx -> \frac {cx}{c}$ so we get
$[a,b] -> [\frac {a}{c},\frac {b}{c}]$

putting this all together, we see that
$\frac {1}{c} \int_{a/c}^{b/c}f(cx)dx=\int_{a}^{b}f(x)dx$

Q.E.D (sort of)

So.....why am i getting $\frac {1}{c}$ when i "should" be getting just $c$???