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Math Help - Area Great area

  1. #1
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    Exclamation Area Great area

    Show that teh exact area of the region boundd by the graph of y= 2sqrt(x^2+1) the coordinate axes and the line x=1, is sqrt2 + ln(sqrt(2)+1)
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  2. #2
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    Hello, Bust20001

    We need Trig substitution for this one. . .


    Show that the exact area of the region bounded by the graph of y\:= \:2\sqrt{x^2+1},
    the coordinate axes and the line x=1 is: . \sqrt2 + \ln(1 + \sqrt2)
    A sketch is always welcome . . .
    Code:
              |       *
              |
              |      *
              |     *|
              |   *::|
              *::::::|
              |::::::|
              |::::::|
          ----+------+----
              0      1
    A \;=\;2\int^1_0\left(x^2+1\right)^{\frac{1}{2}}dx

    Let: x\:=\:\tan\theta\quad\Rightarrow\quad dx \:=\:\sec^2\!\theta\,d\theta

    Substitute: . A\;=\;2\int\sec\theta(\sec^2\!\theta\,d\theta)\;=\  ;2\int\sec^3\!\theta\,d\theta\;=\; \sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|

    Back-substitute: . A \;=\;x\sqrt{x^2+1} + <br />
\ln\left(x+\sqrt{x^2+1}\right)\,\bigg]^1_0

    . . = \;\left[1\cdot\sqrt{2} + \ln\left(1 + \sqrt{2}\right)\right]- \left[0\cdot\sqrt{1} + \ln(0+1)\right]

    . . = \;\sqrt{2} +\ln\left(1 + \sqrt{2}\right)

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