# Integral substitution fallacy

1. Mar 22, 2015

### Wminus

Let $x'=1/u' \Rightarrow dx' = \frac{-1}{u'^2} du'$. Then the integral $\int_{x_0}^{x} x' dx'$ turns into $\int_{1/u_0}^{1/u} \frac{-1}{u'^3} du'$.

Here comes the fallacy: $\int_{1/u_0}^{1/u} \frac{-1}{u'^3} du' = [\frac{1}{2} \frac{1}{u'^2}]_{1/u_0}^{1/u} = \frac{1}{2} (u^2-u_0^2)$. Here I just substituted $1/u$ and $1/u_0$ into $u'$, and I end up with getting something that is obviously wrong since $\frac{1}{2} (u^2-u_0^2) = \frac{1}{2}(1/x^2 - 1/x_0^2) \neq \frac{1}{2}(x^2-x_0^2)$.

2. Mar 22, 2015

### Staff: Mentor

Two things:
1. It's very confusing to see the primes on your variables. I don't think x' denotes the derivative in your example. I could be mistaken, but if I'm not, using x' instead of x adds needless confusion.
2. It's bad practice to use the same variable for a limit of integration and as the dummy variable for integration.
Let x = 1/u, so dx = -du/u2

Also, if x = a, then u = 1/a, and if x = b, then u = 1/b.
Then $\int_a^b x dx = -\int_{1/a}^{1/b} (1/u)(1/u^2)du$
$= -\int_{1/a}^{1/b} u^{-3}du = + (1/2) \left. u^{-2} \right|_{1/a}^{1/b} = (1/2) (b^2 - a^2)$

If you do the integration with no substitution, you get
$\int_a^b x dx = \left. (1/2) x^2 \right|_a^b = (1/2) (b^2 - a^2)$

3. Mar 22, 2015

### mathman

Using u'= 1/x' leads to a u' integral with limits 1/x and 1/x_0.

4. Mar 22, 2015

### Wminus

aah, that's right. I screwed up while setting up the limits. Thank you :)