anonymous
  • anonymous
sketch and then calculate the area of the region bounded by the graphs of x=1, y=x+1 and y=1/sqrt(x^2+1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Is it possible to sketch the graph \(y = \frac{1}{\sqrt{x^2 + 1}}\)? \[\]
anonymous
  • anonymous
I think I would use wolfram alpha to sketch the graph.
anonymous
  • anonymous
Such a complicated Graph to sketch all by yourself.

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anonymous
  • anonymous
Okay I got it We need to fins the intersection of \(\frac{1}{\sqrt{x^2+1}}=y \) and y = x + 1 \[x + 1 = \frac{1}{\sqrt{x^2+1}} \implies (x+1)^2 (x^2+1) = 1\] To do it quickly I will use wolfram alpha but you should solve on your own. BTW from the graph it is clearly visible that the point is x =0 and y=1 So here is the equation you should use to get the area of the region required \[\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx\]
anonymous
  • anonymous
I did all the talking and solving :-/
anonymous
  • anonymous
sorry somthing came up
anonymous
  • anonymous
i appreciate your help
anonymous
  • anonymous
thats what i tried to do
anonymous
  • anonymous
So, did you get it? I mean the answer did you get the answer?
anonymous
  • anonymous
i got (x^2/2)+x - Lnsec theta+ tan theta
anonymous
  • anonymous
.618626 area?
anonymous
  • anonymous
\[\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx = \left[ \frac{x^2}{2} + x \right]_0^1 - \left[ \tan^{-1}x \right]_0^1 \] I think this is what you are supposed to get.
anonymous
  • anonymous
\[\frac{1}{2} + 1 - \frac{\pi}{4}\]
anonymous
  • anonymous
I am getting .714... something
anonymous
  • anonymous
the second part i got LN(sec theta + tan theta) for integral
anonymous
  • anonymous
did i do it wrong?
anonymous
  • anonymous
Yeah I don't seem to get ln(sec theta) anywhere. You should recheck your solution and I will check mine. Maybe I am wrong.
anonymous
  • anonymous
i let x = tan theta and du = sec^2 theta
anonymous
  • anonymous
did it this way http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x%5E2%2B1%29
anonymous
  • anonymous
Umm Lammy I think \[\frac{1}{\sqrt{x^2+1}} = \tan^{-1}x\] I am not sure. I am gonna apply same substitution as yours \(dx = d\theta \sec^2\theta\) \[\frac{1}{\sqrt{\tan^2\theta + 1}}\times \sec^2\theta d\theta = \sec \theta d\theta \] I think you are right I was wrong, sorry for misleading you
anonymous
  • anonymous
so integral of sec theta = LN abs(sec theta + tan theta) right?
anonymous
  • anonymous
no no you were helping =)
anonymous
  • anonymous
then i just subtract and will get the answer right?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
do you think you have time to help me with on more? befor you go to sleep?
anonymous
  • anonymous
hahah
anonymous
  • anonymous
yeah sure, post it :-D
anonymous
  • anonymous
just posted it
anonymous
  • anonymous
=)

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