Q.Evaluate $\displaystyle\int_{0}^{2} (3x^{2} - 2x + 1)\,dx$.

Integrate term by term using linearity and the power rule, then evaluate from $0$ to $2$.

$\big[x^3 - x^2 + x\big]_0^2 = (8-4+2)-0 = 6$.

Use linearity to integrate each term separately (§2).

Antiderivative.

Take $F(x) = x^3 - x^2 + x$.

Evaluate at the limits.

Check (dual-solve): evaluate the three definite integrals separately and add. $\displaystyle\int_0^2 3x^2\,dx = \big[x^3\big]_0^2 = 8$; $\displaystyle\int_0^2 2x\,dx = \big[x^2\big]_0^2 = 4$; $\displaystyle\int_0^2 1\,dx = \big[x\big]_0^2 = 2$. Combining with the signs of the integrand: $8 - 4 + 2 = 6$, agreeing with the direct evaluation.

A common slip is to forget that the coefficient and the power rule interact: $\int 3x^2\,dx = 3\cdot\tfrac{x^3}{3} = x^3$, not $3x^3$. Divide by the new power after raising it. Also, since the lower limit is $0$ here every term vanishes there — but do not assume that in general; always substitute the lower limit explicitly.