- #1
bigevil
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Homework Statement
Find the arc length of [tex]y=\sqrt{x}[/tex] from x=0 to x=2.
The Attempt at a Solution
I don't know, this is a nastier integral than it looks. From the substitutions,
[tex]s = \int_0^2 \sqrt{1 + \frac{1}{4x}} dx[/tex]. From doing this over and over again I already know the answer will have square roots and logarithms but I keep getting it wrong!
I've tried two methods so far, both end with evaluating [tex]\int cosec^3 x dx[/tex], but with different limits.
Heres one: [tex]u = \sqrt{1 + \frac{1}{4x}}[/tex]. Then, I get [tex]s = \int_1^{\sqrt{9/8}} \frac{u^2}{- 2(u^2-1)^2} du[/tex]
Substitute [tex]u = sec \theta[/tex], [tex]s = \int_{0}^{sec^{-1} \sqrt{(9/8)}} - cosec^3 \theta d\theta = \frac{1}{8}(3\sqrt{2}... [/tex] which has the wrong form.
([tex]2 \int cosec^3 x dx = - cot\theta cosec\theta - ln|cot\theta + cosec\theta| + c[/tex] by integration by parts.)
The answer is [tex]\frac{1}{2} (3\sqrt{2} + ln (1 + \sqrt{2}) [/tex] or about 2.56. I'm quite sure my general method is correct (the square roots and logarithms are consistent with the ansewr), but there's a problem with the limits.
The other one is to substitute a trig expression directly (ie [tex]\frac{1}{4x} = tan^2 \theta[/tex], with different limits.
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