Significance of $\int_a^bf(x)dx=(b-a)\int_0^1f\{(b-a)x+a\}dx$ I learnt this property of definite integrals-

$$\int_a^bf(x)dx=(b-a)\int_0^1f\{(b-a)x+a\}dx$$

Now, is there any nice geometrical interpretation of this property (maybe using areas) that would give me a more intuitive feeling of this?  (I mean, can this transformation from LHS to RHS be shown by a transformation of area?)
Also, are there any problems (questions) where this property would prove remarkably useful and reduce a given definite integral in a more simpler form?
Thanks for any help!
 A: Say two people measure a field that starts a mile from town, extends to West/East from smallville to bigtown, and has a South edge that is straight, and a North edge described (in miles) by $f(x)$ where $x$ is the distance from smallville (again measured in miles).
Person A has an West-East measuring tool that in fact starts at zero at smallville, and is calibrated in miles-to-the-East; she has an easy time, just integrating $f(x)$ from smallville (a) to bigtown, where $x$ is the reading on her measuring tool as she walks along.
Person B has an East-West measuring tool that is linear but it needs to be calibrated.  He calibrates it by going to the West edge of the field and setting the zero point to that value, and then going to the East edge and setting the scale so that the East edge is at $1$.   He then has to figure out, as he walks from West to East, what the N/S height of the field is -- in terms of his West-East measurement $x$, the value he will get is 
$$
f\left((b-a)x + a\right)$$
because the distance from smallville is $(b-a)x + a$.  So he does this integration as he walks along. But that integration comes up with a different answer because he has set his unit of West-East distance such that the edges of the field are $1$ apart, whole they are really $b-a$ apart. To correct for this, he needs to multiply the naive area he has calculated by $b-a$, and that completes the formula you have.
In general, any monotonic change in variables of integration can be looked at as applying a different (perhaps in a non-linear way) scale or tool in measuring the "x" distance. So for example, if he had a tool that measured the square of the distance from smallville, and $f(x)$ happened to be contrived such that the integral was easier using the square of the distance, then this tells you how to make a "change in variables" to get an equivalent integral which might be easier (or harder) to calculate. 
