Y = (1/4) x^2 - (1/2) lnx ... over the interval (1, 7e) ... what is the arc length?

Question

Mathematics

Dixon

Today, 06:21

Y = (1/4) x^2 - (1/2) lnx ... over the interval (1, 7e) ... what is the arc length?

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Roberto Vega · Answer 1

So, f[x] = 1/4x^2 - 1/2Ln (x)

thus f'[x] = 1/4*2x - 1/2 * (1/x) = x/2 - 1/2x

thus f'[x]^2 = (x^2) / 4 - 2 * (x/2) * (1/2x) + 1 / (4x^2) = (x^2) / 4 - 1/2 + 1 / (4x^2)

thus f'[x]^2 + 1 = (x^2) / 4 + 1/2 + 1 / (4x^2) = (x/2 + 1/2x) ^2

thus Sqrt[ ... ] = (x/2 + 1/2x)