The question asks to evaluate the definite integral:

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

Step 1: Find the indefinite integral

We will integrate each term separately.

• For $2x^3$, the integral is: $\int 2x^3 \, dx = \frac{2x^4}{4} = \frac{x^4}{2}$
• For $-3x^2$, the integral is: $\int -3x^2 \, dx = -\frac{3x^3}{3} = -x^3$
• For $x$, the integral is: $\int x \, dx = \frac{x^2}{2}$

So, the indefinite integral of the expression is: $\frac{x^4}{2} - x^3 + \frac{x^2}{2}$

Step 2: Evaluate the definite integral

Now, we evaluate the expression from 0 to 1.

$\left[ \frac{x^4}{2} - x^3 + \frac{x^2}{2} \right]_0^1$

Substituting $x = 1$: $\frac{1^4}{2} - 1^3 + \frac{1^2}{2} = \frac{1}{2} - 1 + \frac{1}{2} = 0$

Substituting $x = 0$: $\frac{0^4}{2} - 0^3 + \frac{0^2}{2} = 0$

Thus, the value of the definite integral is: $0 - 0 = 0$

Final Answer:

The correct answer is 0, which corresponds to Response 3 in the image.

Would you like more details or have any other questions? Here are five related questions:

1. How do you calculate definite integrals using the Fundamental Theorem of Calculus?
2. What are common mistakes to avoid when integrating polynomials?
3. How would the result change if the limits of integration were different?
4. What are the applications of definite integrals in real-life scenarios?
5. How do you evaluate definite integrals involving trigonometric functions?

Tip: When evaluating definite integrals, always ensure that you simplify and check the substitution carefully to avoid calculation errors.