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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)

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

  2. anonymous
    • 4 years ago
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    I think I would use wolfram alpha to sketch the graph.

  3. anonymous
    • 4 years ago
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    Such a complicated Graph to sketch all by yourself.

  4. anonymous
    • 4 years ago
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    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 \[\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx\]

  5. anonymous
    • 4 years ago
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    I did all the talking and solving :-/

  6. anonymous
    • 4 years ago
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    sorry somthing came up

  7. anonymous
    • 4 years ago
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    i appreciate your help

  8. anonymous
    • 4 years ago
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    thats what i tried to do

  9. anonymous
    • 4 years ago
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    So, did you get it? I mean the answer did you get the answer?

  10. anonymous
    • 4 years ago
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    i got (x^2/2)+x - Lnsec theta+ tan theta

  11. anonymous
    • 4 years ago
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    .618626 area?

  12. anonymous
    • 4 years ago
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    \[\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx = \left[ \frac{x^2}{2} + x \right]_0^1 - \left[ \tan^{-1}x \right]_0^1 \] I think this is what you are supposed to get.

  13. anonymous
    • 4 years ago
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    \[\frac{1}{2} + 1 - \frac{\pi}{4}\]

  14. anonymous
    • 4 years ago
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    I am getting .714... something

  15. anonymous
    • 4 years ago
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    the second part i got LN(sec theta + tan theta) for integral

  16. anonymous
    • 4 years ago
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    did i do it wrong?

  17. anonymous
    • 4 years ago
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    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.

  18. anonymous
    • 4 years ago
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    i let x = tan theta and du = sec^2 theta

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

  20. anonymous
    • 4 years ago
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    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

  21. anonymous
    • 4 years ago
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    so integral of sec theta = LN abs(sec theta + tan theta) right?

  22. anonymous
    • 4 years ago
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    no no you were helping =)

  23. anonymous
    • 4 years ago
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    then i just subtract and will get the answer right?

  24. anonymous
    • 4 years ago
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    yeah

  25. anonymous
    • 4 years ago
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    do you think you have time to help me with on more? befor you go to sleep?

  26. anonymous
    • 4 years ago
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    hahah

  27. anonymous
    • 4 years ago
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    yeah sure, post it :-D

  28. anonymous
    • 4 years ago
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    just posted it

  29. anonymous
    • 4 years ago
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    =)

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