Parametric Arc Length Integral Setup

Question

Given the parametric equations $$x(t)=\ln(t)$$ and $$y(t)=\sqrt{t}$$ for $$t\in [1,e^2]$$, which of the following integrals correctly represents the arc length of the curve?
 
| t    | x(t)=ln(t) | y(t)=√t |
|------|------------|---------|
| 1    | 0          | 1       |
| e    | 1          | √e      |
| e^2  | 2          | e       |

Accepted Answer

$$\int_{1}^{e^2} \sqrt{\frac{1}{t^2}+\frac{1}{4*t}}\,dt$$

Answer

$$\int_{1}^{e^2} \left(\frac{1}{t}+\frac{1}{2\sqrt{t}}\right)dt$$

Answer

$$\frac{1}{2}\int_{1}^{e^2} \sqrt{\frac{1}{t^2}+\frac{1}{4*t}}\,dt$$

Answer

$$\int_{1}^{e^2} \sqrt{\frac{1}{t}+\frac{1}{4*t}}\,dt$$

Parametric Arc Length Integral Setup

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