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I have the integral expression for the arc length: $$ \int \sqrt{1 + \frac{1}{4x}\,}\,\mathrm{d}x $$

and need to approximate the arc length of the curve: $y = $$ \sqrt{x}$ in between x = 0 and x = 4.

Should I be using the trapezoidal rule or Simpson's rule for this? And how can I begin to apply it?

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The formula for arc length of $y = \sqrt{x}$ from $x = 0$ to $x=4$ is $$ \int_{0}^{4} \sqrt{ 1 + \left( \frac{dy}{dx} \right)^2 } dx$$ where here $\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$, squaring this results in $\left( \frac{dy}{dx} \right)^2 = \frac{1}{4x}$. Hence our task is to integrate $$\int_{0}^{4} \sqrt{1 + \frac{1}{4x}} dx$$ Let $$u = \sqrt{ 1 + \frac{1}{4x} } \Rightarrow u^2 = 1 + \frac{1}{4x} \Rightarrow 2udu = \frac{-1}{4x^2}dx$$ From that we also obtain that $x = \frac{1}{4(u^2 - 1)}$ so that $x^2 = \frac{1}{16(u^2 - 1)^2}$, then $\frac{-1}{4x^2} = -4(u^2-1)^2$, so that we get $$dx = \frac{-u}{2(u^2 -1)^2}du$$ Hence, with our sub, we get the integral $$\int_{x=0}^{x=4} \frac{-u^2}{2(u^2 - 1)^2} du$$ Now I would do another sub, letting $u = \sec(\theta)$ so that $du = \sec(\theta)\tan(\theta)d\theta$, and we get with Phythagoras' Theorem that $$\int_{x=0}^{x=4} \dfrac{-\sec^2(\theta)}{2\tan^4(\theta)} \sec(\theta)\tan(\theta)d\theta = -\dfrac{1}{2} \int_{x=0}^{x=4} \dfrac{\sec^3(\theta)}{\tan^3(\theta)} d\theta = -\dfrac{1}{2} \int_{x=0}^{x=4} \csc^3(\theta)d\theta,$$ This can be solved via integration by parts and is definitely done in detail somewhere on the internet. You should arrive at the answer $$= \dfrac{-1}{2} \left( \dfrac{-\csc(\theta)\cot(\theta) + ln|\csc(\theta) - \cot(\theta)|}{2} \right)$$ I will leave it to you to do the back substitutions and plug in the bounds. Also, I feel like I pulled out a sledge hammer to solve this, I'm sure someone will post a more elegant solution.

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  • $\begingroup$ I really appreciate you taking the time to answer. Before I asked the question, I came to your answer as well but I feel as though there's an easier way to do it as we haven't learned any integration by parts, only Simpson's rule and trapezoidal rule. I feel these are the things being tested on, I'm just not sure how to go about applying any one of them to solving the arc length. $\endgroup$ – Broadsword93 May 9 '17 at 3:16
  • $\begingroup$ For example, with the trapezoidal rule [using this video youtube.com/watch?v=1p0NHR5w0Lc] , I'm worried that I might be finding the area under the curve rather than the arc length. $\endgroup$ – Broadsword93 May 9 '17 at 3:17
  • $\begingroup$ And with Simpson's rule, we aren't given any n value to use in the formula, so I'm guessing I have to make up my own number of intervals, which would probably be 4. Not sure if I'm on the right track with my line of thinking. $\endgroup$ – Broadsword93 May 9 '17 at 3:19
  • $\begingroup$ @user411697 you don't solve arc length using those rules, you approximate it. Those are numerical integration techniques; asking how to do numerical integration is a totally separate question than asking to find the arc length of a curve (which implies you want an exact answer) $\endgroup$ – Brevan Ellefsen May 9 '17 at 3:26
  • $\begingroup$ The area under the curve of $\int_0^4 \sqrt{ 1 + \frac{1}{4x}}$, is identically the same as the length of the curve of $\sqrt{x}$ from $x=0$ to $x=4$. That's what that arc length formula is stating. What you're after with Simpson's or Trapezoidal rule is a mere approximation. The work I showed if carried out would give an exact solution. It would be far more computationally easier to use Simpson or Trapezodial. As far as using Simpson's for your comment the larger the $n$ the closer you'll obtain to the exact solution. $\endgroup$ – Dragonite May 9 '17 at 3:27

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