Answer
$\frac{1}{3}$
Work Step by Step
$\int^1_0 (2t-1)^2dt$
$\int^1_0 (4t^2 -4t+1)dt$, Rewrite
$\frac{4}{3}t^3 -2t^2 +t$, integrate
${\frac{4}{3}t^3 -2t^2 +t}]^1_0$, Definite Integral Form
$(\frac{4}{3}-2+1)$, Definite Integral
$\frac{1}{3}$, simplify
