The question is asking for the value of the definite integral:

$\int_0^1 \left( 3x^2 - 2x + 1 \right) dx$

Let me calculate it step by step:

Step 1: Find the antiderivative

The integral of $3x^2 - 2x + 1$ is computed term by term: $\int \left( 3x^2 - 2x + 1 \right) dx = x^3 - x^2 + x + C$

So, the antiderivative is: $F(x) = x^3 - x^2 + x$

Step 2: Evaluate the definite integral

Now we evaluate $F(x)$ at the bounds $x = 1$ and $x = 0$: $\int_0^1 \left( 3x^2 - 2x + 1 \right) dx = F(1) - F(0)$

Compute $F(1)$:

$F(1) = (1)^3 - (1)^2 + (1) = 1 - 1 + 1 = 1$

Compute $F(0)$:

$F(0) = (0)^3 - (0)^2 + (0) = 0$

Final result:

$\int_0^1 \left( 3x^2 - 2x + 1 \right) dx = 1 - 0 = 1$

Final Answer:

The value of the definite integral is: $\boxed{1}$

Would you like me to explain any part of the solution in more detail or provide additional examples?

Follow-Up Questions:

1. What is the general process for solving definite integrals?
2. How do you handle definite integrals where the bounds are reversed (e.g., $\int_b^a$)?
3. Can definite integrals result in negative values? Why or why not?
4. How does the Fundamental Theorem of Calculus relate to definite integrals?
5. What happens when integrating higher-degree polynomials?

Tip:

When solving definite integrals, always ensure that you correctly evaluate the antiderivative at the given bounds.