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## Homework Statement

Find the arclength of $$r(t) = <{t}^2/2,{t}^3/3>$$ from t=-1 to t=1

## Homework Equations

I have used this equation for arclength $$\int_{-1}^{1}|{r}'(t)|dt$$

## The Attempt at a Solution

After integrating (using u substitution) I have the solution

$$\frac{1}{3}({t^2+1})^\frac{3}{2}$$

**evaluated from t = -1 to t = 1.**Of course, this yields 0! Which doesn't make any sense for arclength.

I know that the problem is somewhere in my integration step because in mathematica both the ArcLength[ $$\int_{-1}^{1}|{r}'(t)|dt$$] function and the Integrate[$$\sqrt{t^2+t^4}$$] give the same result $$\frac{2}{3}(-1+2\sqrt{2})$$, which seems to be the solution. What am I doing wrong?