- 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)

- jamiebookeater

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- anonymous

Is it possible to sketch the graph \(y = \frac{1}{\sqrt{x^2 + 1}}\)? \[\]

- anonymous

I think I would use wolfram alpha to sketch the graph.

- anonymous

Such a complicated Graph to sketch all by yourself.

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- 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

I did all the talking and solving :-/

- anonymous

sorry somthing came up

- anonymous

i appreciate your help

- anonymous

thats what i tried to do

- anonymous

So, did you get it? I mean the answer did you get the answer?

- anonymous

i got (x^2/2)+x - Lnsec theta+ tan theta

- anonymous

.618626 area?

- 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

\[\frac{1}{2} + 1 - \frac{\pi}{4}\]

- anonymous

I am getting .714... something

- anonymous

the second part i got LN(sec theta + tan theta) for integral

- anonymous

did i do it wrong?

- 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

i let x = tan theta and du = sec^2 theta

- anonymous

did it this way http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt%28x%5E2%2B1%29

- 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

so integral of sec theta = LN abs(sec theta + tan theta) right?

- anonymous

no no you were helping =)

- anonymous

then i just subtract and will get the answer right?

- anonymous

yeah

- anonymous

do you think you have time to help me with on more? befor you go to sleep?

- anonymous

hahah

- anonymous

yeah sure, post it :-D

- anonymous

just posted it

- anonymous

=)

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