1. The problem asks to find the value of the definite integral $$\int_0^2 (x^2 - 2x + 1) \, dx$$. 2. The formula for the definite integral of a function $f(x)$ from $a$ to $b$ is: $$\int_a^b f(x) \, dx = F(b) - F(a)$$ where $F(x)$ is the antiderivative of $f(x)$. 3. First, find the antiderivative of the integrand $x^2 - 2x + 1$: $$\int (x^2 - 2x + 1) \, dx = \frac{x^3}{3} - x^2 + x + C$$ 4. Evaluate the antiderivative at the upper limit $x=2$: $$F(2) = \frac{2^3}{3} - 2^2 + 2 = \frac{8}{3} - 4 + 2 = \frac{8}{3} - 2 = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}$$ 5. Evaluate the antiderivative at the lower limit $x=0$: $$F(0) = \frac{0^3}{3} - 0^2 + 0 = 0$$ 6. Calculate the definite integral value: $$\int_0^2 (x^2 - 2x + 1) \, dx = F(2) - F(0) = \frac{2}{3} - 0 = \frac{2}{3}$$ 7. Therefore, the value of the integral is $\frac{2}{3}$. The correct answer is C. 2/3.