To evaluate the definite integral

$\int_{-1}^2 (3x^2 - 2x + 3) \, \mathrm{d}x,$

we will find the antiderivative of the function $3x^2 - 2x + 3$ and then apply the Fundamental Theorem of Calculus.

Step 1: Find the antiderivative

The antiderivative of $3x^2 - 2x + 3$ is calculated term by term:

• For $3x^2$, the antiderivative is $\frac{3x^3}{3} = x^3$.
• For $-2x$, the antiderivative is $-\frac{2x^2}{2} = -x^2$.
• For $3$, the antiderivative is $3x$.

Thus, the antiderivative is:

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

Step 2: Apply the limits

Using the limits $-1$ and $2$, we compute:

$\int_{-1}^2 (3x^2 - 2x + 3) \, \mathrm{d}x = F(2) - F(-1),$

where: $F(2) = (2)^3 - (2)^2 + 3(2) = 8 - 4 + 6 = 10,$ $F(-1) = (-1)^3 - (-1)^2 + 3(-1) = -1 - 1 - 3 = -5.$

Step 3: Final calculation

$F(2) - F(-1) = 10 - (-5) = 10 + 5 = 15.$

Final Answer:

$\int_{-1}^2 (3x^2 - 2x + 3) \, \mathrm{d}x = 15.$

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Would you like further clarification on any steps or additional examples? Here are some related questions:

1. How is the Fundamental Theorem of Calculus applied in general cases?
2. What are some practical applications of definite integrals?
3. Can you show how to compute the integral using numerical methods?
4. How would the result change if the limits were swapped?
5. What does the definite integral represent geometrically in this context?

Tip: Always simplify the antiderivative step-by-step to avoid calculation errors, especially when multiple terms are involved.