Show that: $ \int_{1}^{3}\int_{0}^{x-1} f_{x}(x,y)dydx = -\frac{4}{3}$

Question

Let f(x,y) be a differentiable function satisfying $f(3,y-1) = (y-1)(y-3) + f(y,y-1)$

Show that:

$$ \int_{1}^{3}\int_{0}^{x-1} f_{x}(x,y)dydx = -\frac{4}{3}$$

I'm not sure how to even begin this question. How do I turn the first equation into an form which I can integrate?

Answer 1

For anyone who doesn't get it, switch the order of integration as so:

$(1)$ $ 0 \le y \le x-1 \\$
$(2)$ $1 \le x \le 3 $

From (1) we can use our x-max to get our y-bounds

$ 0 \le y \le 2 \\$

and then we can use our identities to find out x bound.

$y + 1 \le x \le 3 \\$

The rest is trivial