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anonymous
 4 years ago
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)
anonymous
 4 years ago
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)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is it possible to sketch the graph \(y = \frac{1}{\sqrt{x^2 + 1}}\)? \[\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I would use wolfram alpha to sketch the graph.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Such a complicated Graph to sketch all by yourself.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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
 4 years ago
Best ResponseYou've already chosen the best response.0I did all the talking and solving :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry somthing came up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i appreciate your help

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats what i tried to do

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, did you get it? I mean the answer did you get the answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got (x^2/2)+x  Lnsec theta+ tan theta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_0^{1} (x+1)dx\int_0^1 \frac{1}{\sqrt{x^2+1}}dx = \left[ \frac{x^2}{2} + x \right]_0^1  \left[ \tan^{1}x \right]_0^1 \] I think this is what you are supposed to get.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2} + 1  \frac{\pi}{4}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am getting .714... something

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the second part i got LN(sec theta + tan theta) for integral

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah 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
 4 years ago
Best ResponseYou've already chosen the best response.0i let x = tan theta and du = sec^2 theta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did it this way http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x%5E2%2B1%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Umm 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
 4 years ago
Best ResponseYou've already chosen the best response.0so integral of sec theta = LN abs(sec theta + tan theta) right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no no you were helping =)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then i just subtract and will get the answer right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you think you have time to help me with on more? befor you go to sleep?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah sure, post it :D
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