Arc of curve of function - Find the minimum length 
This is the arc of the curve:
$$y = x^3 + x^2 - \frac{29x}{2} + 1 \\ t > 0 \\ x \in [t,t+1]$$
Find $t$ for which the length of the arc of the curve is minimum.

Should I use $ \int {\sqrt{1 + [f'(x)]^2}} dx$ ?
Thank you very, but very much!
 A: Yes, one should use the formula. Note that the integral is from $t$ to $t+1$. 
Then differentiate, using the Fundamental Theorem of Calculus, and proceed as usual. After setting the derivative equal to $0$, be careful when finding square roots.
A: doing André Nicolas say, we have:
$$f'(x)=3x^2+2x-\frac{29}{2}$$
and
$$g(t)=\int_t^{t+1}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx$$ $$g(t)=\int_0^{t+1}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx-\int_0^{t}{\sqrt{1 + \left(3x^2 + 2x - \frac{29}{2}\right)^2}} dx$$
now we calculate $g'(t)$:
$$g'(t)=\sqrt{1 + \left(3(t+1)^2 + 2(t+1) - \frac{29}{2}\right)^2}-\sqrt{1 + \left(3t^2 + 2t - \frac{29}{2}\right)^2}=0$$
$$1 + \left(3(t+1)^2 + 2(t+1) - \frac{29}{2}\right)^2-1-\left(3t^2 + 2t - \frac{29}{2}\right)^2=0$$
$$\left(3t^2+2t-\frac{29}{2}+6t+5\right)^2-\left(3t^2+2t-\frac{29}{2}\right)^2=0$$
$$2(6t+5)(3t^2+2t-\frac{29}{2})+(6t+5)^2=0$$
if $t$ distinc to $-\frac{5}{6}$, we factorizing $(6t+5)$:
$$6t^2+4t-29+6t+5=0$$
finally we have:
$$6t^2+10t-24=0$$
wich have the solutions: $t=-\frac{16}{12}$ and $t=3$
Finally the result is:
$$t=3$$
