# unit of integration vs unit of boundaries

• Oct 12th 2009, 07:41 PM
chug1
unit of integration vs unit of boundaries
Hi
Was just wondering:
In the expression
$\int_{a}^{b}f(x)dx$
it is assumed that a and b are values of x.
Yet when we do u-substitution, to get
$\int_{a}^{b}g(u)du$
we are still assuming that a and b are value of x (the boundaries of the area we are finding, as it were).
But really, shouldn't they also be converted to the corresponding values of u? The unit of integration is now u and everything else in the expression is now in terms of u, yet these boundaries are still referring to values of x. To change f(x) -> g(u) we needed some function mapping x to u, so we must also be able to map the boundaries to u as well.

The conventional method when using u-ubstitution to help with integration involves re-converting the final expression (e.g. 4u^2 +2u +c) back to x, and then substituting in the boundary values of x.
If the boundary values are converted from x-values to u-values, this reversal wouldn't be required before the final evaluation.

While I admit that it is probably just easier to do it the conventional way, my question is whether or not the procedure outlined above is actually correct.

Cheers
• Oct 12th 2009, 08:34 PM
Chris L T521
Quote:

Originally Posted by chug1
Hi
Was just wondering:
In the expression
$\int_{a}^{b}f(x)dx$
it is assumed that a and b are values of x.
Yet when we do u-substitution, to get
$\int_{a}^{b}g(u)du$
we are still assuming that a and b are value of x (the boundaries of the area we are finding, as it were).
But really, shouldn't they also be converted to the corresponding values of u? The unit of integration is now u and everything else in the expression is now in terms of u, yet these boundaries are still referring to values of x. To change f(x) -> g(u) we needed some function mapping x to u, so we must also be able to map the boundaries to u as well.

The conventional method when using u-ubstitution to help with integration involves re-converting the final expression (e.g. 4u^2 +2u +c) back to x, and then substituting in the boundary values of x.
If the boundary values are converted from x-values to u-values, this reversal wouldn't be required before the final evaluation.

While I admit that it is probably just easier to do it the conventional way, my question is whether or not the procedure outlined above is actually correct.

Cheers

You're ideology is correct.

You can actually convert the boundaries of x to boundaries of u!

If $u=h\!\left(x\right)$ is your substitution, then your new limits of integration (i.e. the boundaries) are $a_u=h\!\left(a\right)$ and $b_u=h\!\left(b\right)$.

So, $\int_a^bf\!\left(x\right)\,dx\xrightarrow{u=h\!\le ft(x\right)}{}\int_{a_u}^{b_u}g\!\left(u\right)\,d u$.

Once this is done, you don't have to convert back to x!
• Oct 12th 2009, 08:35 PM
Jose27
Quote:

Originally Posted by chug1
Hi
Was just wondering:
In the expression
$\int_{a}^{b}f(x)dx$
it is assumed that a and b are values of x.
Yet when we do u-substitution, to get
$\int_{a}^{b}g(u)du$
we are still assuming that a and b are value of x (the boundaries of the area we are finding, as it were).
But really, shouldn't they also be converted to the corresponding values of u? The unit of integration is now u and everything else in the expression is now in terms of u, yet these boundaries are still referring to values of x. To change f(x) -> g(u) we needed some function mapping x to u, so we must also be able to map the boundaries to u as well.

The conventional method when using u-ubstitution to help with integration involves re-converting the final expression (e.g. 4u^2 +2u +c) back to x, and then substituting in the boundary values of x.
If the boundary values are converted from x-values to u-values, this reversal wouldn't be required before the final evaluation.

While I admit that it is probably just easier to do it the conventional way, my question is whether or not the procedure outlined above is actually correct.

Cheers

Yes, basically you're thinking of the change of variables theorem that says: If you have $f:[a,b] \rightarrow \mathbb{R}$ and a function $g:[c,d] \rightarrow [a,b]$ that is continously differentiable and with a continously diff. inverse then $\int_{g(c)} ^{g(d)} \ f(x)dx = \int_{c} ^{d} \ f(g(u))g'(u)du$
• Oct 12th 2009, 09:14 PM
chug1
Thank you both very much!