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?
Best Answer
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 $I\approx 1.20827$.