Sketch the region which is determined by the inequalities and hence evaluate
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Originally Posted by matty888 Sketch the region * which is determined by the inequalities and hence evaluate Calculate $\displaystyle \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 $.
Last edited by mr fantastic; Mar 21st 2009 at 12:26 AM.
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