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I am very new to calculus and StackExchange so I'm sorry if I make any mistakes. I want to work out the arc length of:

$y = \sqrt{5x} - 2.023, [0.075, 0.58]$.

I have used the definition of a definite integral and got

$\int_{0.075}^{0.58} \sqrt{1+\left(\frac{\sqrt{5}}{2\sqrt{x}}\right)²} dx$ =$\int_{0.075}^{0.58} \sqrt{1+\frac{5}{4x}} dx$

so far which I think is correct. How would I proceed from here? Would I use u-substitution? Any help is appreciated.

Edit: I let $x = u^2$, so I got:

$\int_{0.075}^{0.58} \sqrt{\frac{4u^2+5}{4u^2}} du$

$\sqrt{4u^2} = 2u$, so

$\int_{0.075}^{0.58} \sqrt{{4u^2+5}}$ $du$.

How would I continue from here? Or is this method not correct?

1 Answers1

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I would get a common denominator in the radical then let $u=\sqrt{x}$: $$I=\int_ {\sqrt{0.075}}^{\sqrt{0.58}} \sqrt{4u^2+5} \; du$$ Now, let $u=\frac{\sqrt{5}\tan{t}}{2}$: $$I=\int_{\cdots}^{\cdots} \frac{5}{2}\sec^3{t} \; dt$$ $$I=\frac{5}{4}\left(\ln{\big| \sec{t}+\tan{t} \big |}+\sec{t}\tan{t}\right) \bigg \rvert_{\cdots}^{\cdots}$$ Just find the new bounds then plug them in. You should get $I\approx 1.20827$.

Ty.
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  • Why are the bounds at the start in a radical? I don't understand that? –  Jun 14 '20 at 16:30
  • Look at the substitution you and I made. $u=\sqrt{x}$ and the original bounds correspond to $x$. You have to square root the bounds when making the aforementioned substitution for the integral to be equivalent. – Ty. Jun 14 '20 at 16:32
  • I see, also, would the new bounds be in terms of t? Since before they were in terms of u and before that in terms of x? Sorry if that question is stupid, I'm very new to calculus. –  Jun 14 '20 at 16:42
  • Yes, the final bounds would be in terms of $t$. I put $\cdots$ instead of the bounds because they are quite annoying. To calculate them you would use $u=\frac{\sqrt{5}\tan{t}}{2}$ and plug in the bounds for $u$ to solve for $t$. For example, the upper bound would be $\arctan{\frac{2}{\sqrt{0.58}}{\sqrt{5}}}$. It's not a stupid question – Ty. Jun 14 '20 at 16:44
  • I understand now. I've been trying to learn trigonometric substitution for the past 12 hours and my brain is melting. Thank you so much this really helped me. –  Jun 14 '20 at 16:51