# Let f (x) =x0( u22u+1 )du then find the least value of f (x) in [-1,1].

8 years ago
$\int_0^x \left(u^2+2 u+1\right) \, du=\frac{x^3}{3}+x^2+x$

its a monotonically increasing function.

f'(x)=(x+1)^2=0, x=-1

minima value=-1/3+1-1=-1/3