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Question

Hi, I'm having doubts to how I'm supposed to draw the graph of the triangle, is it just $y=x\tan(\alpha)$ and $y= -x\tan(\alpha)$ and $x=1$ and sketch the region?.

Also for the integral I calculated it to be

$\frac{1}{2} \left[\tan(\alpha) + \frac{\tan^3(\alpha)}{3 }\right]. $

Is the integral calculation correct?

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Andrew Cavagnino
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  • Yes, this looks fine. – user84413 May 19 '15 at 23:27
  • I tried graphing y=xtan(alpha) on a graphing software and it's just dots on the graph, are you sure it's the right graph – Andrew Cavagnino May 20 '15 at 00:23
  • I think so - for example, if $\alpha=\frac{\pi}{4}$, then R is the region bounded above by $y=x$, below by $y=-x$, and on the right by $x=1$. Similarly, $\alpha=\frac{\pi}{3}$ gives the region bounded above by $y=\sqrt{3}x$, bounded below by $y=-\sqrt{3}x$, and on the right by $x=1$. – user84413 May 20 '15 at 00:33

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