Let f (x) = x 0 ( u22u+1 )du then find the least value of f (x) in [-1,1]. -2u+2

Grade 12

1 Answers

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

its a monotonically increasing function.

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

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


Sher Mohammad,
B.tech, IIT Delhi

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