Area Great area

**Bust2000** · March 3rd 2008, 07:26 PM

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)

**Soroban** · March 3rd 2008, 08:54 PM

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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