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JH
Numerade Educator

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Problem 58 Hard Difficulty

Find the area of the region bounded by the given curves.

$ y = \tan x $ , $ y = \tan^2 x $ , $ 0 \le x \le \frac{\pi}{4} $

Answer

$$
\frac{\pi}{4}-1+\ln \sqrt{2}
$$

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Video Transcript

here we'd like to find the area of the region bounded by the curves. Ten x tan squared X for X between zero power for So here's a rough sketch of the graphs and let me explain why So, First of all, we know that can and tan squared or both zero on X zero and they're both won when excess power for now in between for exes between zero pyro for we have numbers. So we have tan and tan square. So we have zero less than our people two ten eggs less than equal to one on the interval. It's your own empire before, and I know that when you take a number between zero and one and you raise it to a higher power, their number itself gets smaller. So since the blue graph is a higher power of the same number ten, it's going to be smaller than the red graph. And so to confirm this, let's go to actual graphing calculator. So just this before the reddest hand, the bluest hand square and from zero to about power, for we see that the red graph is above the blue ref, so that tells us well First of all, we'LL have a formula for the area. So let's write that out. This area is the integral A to be so here. Zero power before of the absolute value of ten eggs Linus Tan square. And by our previous observation, we noticed that tan squared was below or less than or equal to ten eggs. This tells us that tan eggs minus ten square Dex is positive. And we can use this because if so, if this is positive, that means that the absolute value is just itself. So we can write this as in a girl zero power before ten eggs minus ten squared. Now let's go ahead and rewrite this zero power for CNX minus seek and squared X minus one. So we did here was use a path Agron identity to rewrite Tan Square a Sikh and squared minus one. We can evaluate all of these anti derivatives. The derivative of the tangent is natural log of absolute value of C can't. Here we have a minus tan X, and that becomes so we have a double minus their sort of plus X and rn points zero power for So it's going and plug in those end points. So is playing Pi over four first natural log seeking of Piper for his route, too. Since it's positive we can drop the absolute value there and then tangent apart before is one plus x. So plus power for and then when we plug in zero, we have natural log and it's seeking of zero is one tangent of zero zero and then plus zero. And we know that natural log of one zero So we could actually ignore this whole second term. And we're left over with our final answer. Ellen Route too minus one plus pirate for So that's our area, and that's our final answer.