What is the Arc Length of y=\sqrt{x} from x=0 to x=2?

[bigevil]

Homework Statement

Find the arc length of y=\sqrt{x} 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,

s = \int_0^2 \sqrt{1 + \frac{1}{4x}} dx. 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 \int cosec^3 x dx, but with different limits.

Heres one: u = \sqrt{1 + \frac{1}{4x}}. Then, I get s = \int_1^{\sqrt{9/8}} \frac{u^2}{- 2(u^2-1)^2} du

Substitute u = sec \theta, s = \int_{0}^{sec^{-1} \sqrt{(9/8)}} - cosec^3 \theta d\theta = \frac{1}{8}(3\sqrt{2}... which has the wrong form.

(2 \int cosec^3 x dx = - cot\theta cosec\theta - ln|cot\theta + cosec\theta| + c by integration by parts.)

The answer is \frac{1}{2} (3\sqrt{2} + ln (1 + \sqrt{2}) 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 \frac{1}{4x} = tan^2 \theta, with different limits.

 

[Mark44]

Here's something different you might try. Instead of working with y = sqrt(x), you could instead work with y = x^2. The arc length along your curve between x = 0 and x = 2 is exactly the same as the arc length along the curve between x = 0 and x = sqrt(2). If you draw the graphs of y = sqrt(x) and y = x^2, you should be able to convince yourself of what I'm saying.

The integral becomes:
\int_0^{\sqrt{2}} \sqrt{1 + 4x^2} dx

You can use a trig substitution here, with tan \theta = 2x.

 

[bigevil]

Mark, thanks so much, that worked. You meant that since x^2 is the inverse function, it is exactly symmetrical about the y=x, so (2, root-2) maps to (root-2, 2), right?

But I hope to be able to solve the original problem as-is. Just changing functions unfortunately won't reveal what was wrong with what I did above.