integral question for arc length: convert 1 + (x/2-1/(2x))^2 to (x/2+1/(2x))^2

[chemENGstudent]

Now substituting the value of [f '(x)]² with a = 1, b = 2 in the equation (2), we have:

. . . . .\(\displaystyle \begin{align} \displaystyle L\, &=\, \int_1^2\, \sqrt{\strut 1\, +\, \left(\dfrac{1}{2}x\, -\, \dfrac{1}{2x}\right)^2\,}\, dx

\\ \\ &=\, \int_1^2\, \sqrt{\strut \left(\dfrac{x}{2}\, +\, \dfrac{1}{2x}\right)^2\,}\,dx

\\ \\ &=\, \int_1^2\, \left(\dfrac{x}{2}\, +\, \dfrac{1}{2x}\right)\, dx

\\ \\ &=\, \left[\dfrac{x^{1+1}}{2\, \cdot\, 2}\, +\, \dfrac{1}{2}\, \ln(x)\right]_1^2

\\ \\ &=\, \dfrac{3}{4}\, +\, \dfrac{1}{2}\, \ln(2) \end{align}\)

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I don't understand how the 1+ disappeared in the square root and the sign changed.

Please help. Thank you!

 

[chemENGstudent]

Help with Arc Length

Now substituting the value of [f '(x)]² with a = 1, b = 2 in the equation (2), we have:

. . . . .\(\displaystyle \begin{align} \displaystyle L\, &=\, \int_1^2\, \sqrt{\strut 1\, +\, \left(\dfrac{1}{2}x\, -\, \dfrac{1}{2x}\right)^2\,}\, dx

\\ \\ &=\, \int_1^2\, \sqrt{\strut \left(\dfrac{x}{2}\, +\, \dfrac{1}{2x}\right)^2\,}\,dx

\\ \\ &=\, \int_1^2\, \left(\dfrac{x}{2}\, +\, \dfrac{1}{2x}\right)\, dx

\\ \\ &=\, \left[\dfrac{x^{1+1}}{2\, \cdot\, 2}\, +\, \dfrac{1}{2}\, \ln(x)\right]_1^2

\\ \\ &=\, \dfrac{3}{4}\, +\, \dfrac{1}{2}\, \ln(2) \end{align}\)

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I don't understand how the 1+ disappeared in the square root and the sign changed.

Please help. Thank you!

 

[stapel]

Now substituting the value of [f '(x)]² with a = 1, b = 2 in the equation (2), we have:

. . . . .\(\displaystyle \begin{align} \displaystyle L\, &=\, \int_1^2\, \sqrt{\strut 1\, +\, \left(\dfrac{1}{2}x\, -\, \dfrac{1}{2x}\right)^2\,}\, dx

\\ \\ &=\, \int_1^2\, \sqrt{\strut \left(\dfrac{x}{2}\, +\, \dfrac{1}{2x}\right)^2\,}\,dx

\\ \\ &=\, \int_1^2\, \left(\dfrac{x}{2}\, +\, \dfrac{1}{2x}\right)\, dx

\\ \\ &=\, \left[\dfrac{x^{1+1}}{2\, \cdot\, 2}\, +\, \dfrac{1}{2}\, \ln(x)\right]_1^2

\\ \\ &=\, \dfrac{3}{4}\, +\, \dfrac{1}{2}\, \ln(2) \end{align}\)

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I don't understand how the 1+ disappeared in the square root and the sign changed.

What did you get when you applied what you learned back in algebra, and you expanded the integrand and simplified? You started with:

. . . . .\(\displaystyle 1\, +\, \left(\dfrac{1}{2}x\, -\, \dfrac{1}{2x}\right)^2\)

You expanded:

. . . . .\(\displaystyle 1\, +\, \dfrac{1}{4}x^2\, -\, 2\, \cdot\, \left(\dfrac{1}{2}x\right)\, \left(\dfrac{1}{2x}\right)\, +\, \dfrac{1}{4x^2}\)

. . . . .\(\displaystyle 1\, +\, \dfrac{1}{4}x^2\, -\, \dfrac{1}{2}\, +\, \dfrac{1}{4x^2}\)

. . . . .\(\displaystyle \dfrac{1}{4}x^2\, +\, \dfrac{1}{2}\, +\, \dfrac{1}{4x^2}\)

...and... then what?

Please be complete. Thank you!