Question

Make the appropriate substitution so that
\[\int_a^b f(x)~dx=\int_{-1}^1 g(t)~dt\]
where \(a,b\) are distinct and arbitrary real numbers in the domain of \(f\). Good practice in calculus/algebra.

Answer 1

Assume \(f\) is a function such that integrating over these intervals works, etc...

Answer 2

there are infinite options...?

Answer 3

I suppose there are... make the simplest one?

Answer 4

make a bijection of f: v -> w for all integers being part of bipartite sets

Answer 5

let x=t+1 c=a-1 d=b-1

Answer 6

yes?

Answer 7

I seem to have made the mistake of trying to generalize a neat substitution method... Hold on, I'll edit the question.

Answer 8

I suppose we want \(t = g(x)\) where \(g(b) = d\) and \(g(a) = c\).

Are you sure this is always possible?

Answer 9

Regarding the edit: Essentially, you're asked to transform the integral such that you can integrate over an arbitrary interval in one domain, while keeping the interval fixed in another. Does that make sense?

Answer 10

Let \(f(x)=1\), then clearly it isn't always the case that \(b-a=d-c\)

Answer 11

Ever time, I mean

Answer 12

so essentially we are mapping f(x) into [-1,1], isn't that a hilberts hotel problem?

Answer 13

it can only be done if the set is countable

Answer 14

Do you mean
\[\int_a^b f(x)~dx=\int_{-1}^1 g(t)~dt\]
this?

Answer 15

http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel you sure this doesn't apply?

Answer 16

oh

Answer 17

The only way we can make a substitution with the function staying the same, that I can readily think of, would be \(t=x+c\)

Answer 18

Yes @FibonacciChick666 the approach I have in mind has nothing to do with it.

Answer 19

hmm ok, but it is almost like a havyside function I guess, but that has specific scenarios

Answer 20

That would shift the bounds, it wouldn't be able to change the scale of them.

Answer 21

heavyside*