Evaluate the definite integral ${\int }_0^1(3x^2-2x+1)dx$.

Step-by-step solution

1. Find the antiderivative of the integrand: $\int (3x^2-2x+1)dx=x^3-x^2+x+C$.
2. Apply the Fundamental Theorem of Calculus: evaluate the antiderivative at the upper limit $x=1$ and lower limit $x=0$.
3. Substitute $x=1$: $1^3-1^2+1=1-1+1=1$.
4. Substitute $x=0$: $0^3-0^2+0=0$.
5. Subtract the lower limit value from the upper limit value: $1-0=1$.
6. Since the antiderivative includes the constant $C$, it cancels in the subtraction, leaving the result $1$.
7. Re‑check the integration: the term $3x^2$ integrates to $x^3$, the term $-2x$ integrates to $-x^2$, and the constant $1$ integrates to $x$. Hence the correct result is $x^3-x^2+x$ evaluated from 0 to 1, which equals $1-0+0=1$.
8. Therefore the final value of the integral is $\frac{4}{3}$ (after correcting the arithmetic: $x^3=1,-x^2=-1,+x=1$ gives $1-1+1=1$; the integral of $3x^2$ from 0 to 1 is $1$, of $-2x$ is $-1$, and of $1$ is $1$; sum = $1-1+1=1$).

Concept tested: Definite integral evaluation

Published 13 September 2026 · How this solution was prepared and checked