# Thread: Inequalities, double integrals and regions

1. ## Inequalities, double integrals and regions

Sketch the region ­ which is determined by the inequalities and hence evaluate

2. Originally Posted by matty888
Sketch the region * which is determined by the inequalities and hence evaluate

Calculate $\int_{x = -\infty}^0 \int_{y = \frac{1}{x}}^{y = 1 - \frac{1}{x}} \frac{2y}{x^2} \, dy \, dx + \int^{+\infty}_{x = 2} \int_{y = \frac{1}{x}}^{y = 1 - \frac{1}{x}} \frac{2y}{x^2} \, dy \, dx$.