# Double Integral with weird boundaries

Let $A$ be the area of $\Bbb R^2$ restricted by the $y$-axis and $y=e^x, y=2e^{5x}$ and $y=1/x$

Compute the following integral numerically:

$\int\int_A x^2y$ sin$(xy)dxdy$

The numerical part shouldn't be an issue, but I have no idea as to how I'm going to find the boundaries.

-

Make yourself a sketch of the different curves. Next, try to find the intersection points. The graph of $y=\operatorname{e}^x$ meets the $y$-axis when $x=0$, i.e. when $y=\operatorname{e}^0=1$. The graph of $y=2\operatorname{e}^{5x}$ meets the $y$-axis when $x=0$, i.e. when $y=2\operatorname{e}^0 = 2$. So the points $(0,1)$ and $(0,2)$ are vertices of the region.

Next, we want to know when the graph of $y=2\operatorname{e}^{5x}$ meets the graph of $y = \tfrac{1}{x}$. This happens when $2\operatorname{e}^{5x} = \tfrac{1}{x}$. This is highly non-trivial and involves Lambert's $W$-function. However, since this is a numerical problem, you could find the intersection using a numerical method. My favourite is the Newton-Raphson method. Consider $\operatorname{f}(x) := 2\operatorname{e}^{5x} - \tfrac{1}{x}$. We have $\operatorname{f}'(x) = 10\operatorname{e}^{5x}+\tfrac{1}{x^2}$. Then:

$$x_{n+1} := x_n - \frac{\operatorname{f}(x_n)}{\operatorname{f}'(x_n)}$$

Taking $x_0 := 1$ as an intial guess, we get $x_1 \approx 0.8$, $x_2 \approx 0.6$, $x_3 \approx 0.4$, $x_4 \approx 0.27$, $x_5 \approx 0.19$, and so on. By $x_8$ the values have settled down to $x_{\ge 8} \approx 0.1917172714$. Hence the point $(x,y)=(0.191717271,5.21601415)$ is another vertex of the region.

Finally, we need to find where the graph $y=\operatorname{e}^x$ meets the graph $y = \tfrac{1}{x}$. Again, this involves the Lambert $W$-function. Using numerical methods we arrive at $x \approx 0.5671432904$. Hence the point $(x,y)=(0.5671432904,1.763222834)$ is the final vertex of the region.

The region is give approximately by $\operatorname{e}^x \le y \le 2\operatorname{e}^{5x}$ for all $0 \le x < 0.1917172714$. Then $\operatorname{e}^x \le y \le \tfrac{1}{x}$ for all $0.1917172714 \le x \le 0.5671432904$.

-
Thanks! :) That was very helpful! – L1meta Apr 21 '13 at 15:19
@L1meta No worries. I'm glad I could help. – Fly by Night Apr 21 '13 at 15:28