Double Integral -- can you pinpoint the error?

Question

I am trying to integrate a function over the area of a triangle as in the image below, but I get an erroneous result (0). Could you pinpoint to me where I err in my calculations? Your advice will be appreciated:

Answer 1

The indefinite integral of $u^{-2}-1/4$ is $-u^{-1}-u/4$; you don't seem to have substituted $u=2$ and $u=1$ successfully into this.

(And previously the integral of $3x^{-3}$ is $-(3/2)x^{-2}$ not $-x^{-2}$.)

Answer 2

$$ \int_u^2 3 x^{-3} {\rm d}x = -\frac{3}{8} + \frac{3}{2 u^2} $$

Therefore

$$ \int_1^2\int_u^2 3 x^{-3} {\rm d}x{\rm d}u = \int_1^2\left[-\frac{3}{8} + \frac{3}{2 u^2}\right]{\rm d}u = \frac{3}{8} $$

Answer 3

$$\int_{1}^{2}\int_{u}^{2}3x^{-3}dxdu=\int_{1}^{2}-\frac{3}{2}x^{-2}\bigg\vert_{u}^{2}du=\int_{1}^{2}-\frac{3}{2}\left(\frac{1}{4}-u^{-2}\right)du=-\frac{3}{2}\left(\frac{1}{4}u+u^{-1}\bigg\vert_{1}^{2}\right)=-\frac{3}{2}\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{4}-1\right)=\frac{3}{8}$$