To evaluate the definite integral

$\int_{-1}^{2} (x-1)^2 \, dx,$

we proceed step by step:

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Step 1: Expand the expression

The integrand $(x-1)^2$ can be expanded as: $(x-1)^2 = x^2 - 2x + 1.$

So, the integral becomes: $\int_{-1}^{2} (x^2 - 2x + 1) \, dx.$

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Step 2: Integrate term by term

The integral is split into three separate integrals: $\int_{-1}^{2} (x^2 - 2x + 1) \, dx = \int_{-1}^{2} x^2 \, dx - 2 \int_{-1}^{2} x \, dx + \int_{-1}^{2} 1 \, dx.$

1. For $\int x^2 \, dx$: $\int x^2 \, dx = \frac{x^3}{3} + C.$

2. For $\int x \, dx$: $\int x \, dx = \frac{x^2}{2} + C.$

3. For $\int 1 \, dx$: $\int 1 \, dx = x + C.$

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Step 3: Apply the limits

Substitute the limits of integration into the antiderivative.

1. First term: $\left[ \frac{x^3}{3} \right]_{-1}^{2} = \frac{2^3}{3} - \frac{(-1)^3}{3} = \frac{8}{3} - \left(-\frac{1}{3}\right) = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3.$

2. Second term: $-2 \left[ \frac{x^2}{2} \right]_{-1}^{2} = -2 \left( \frac{2^2}{2} - \frac{(-1)^2}{2} \right) = -2 \left( \frac{4}{2} - \frac{1}{2} \right) = -2 \left( 2 - 0.5 \right) = -2 \cdot 1.5 = -3.$

3. Third term: $\left[ x \right]_{-1}^{2} = 2 - (-1) = 2 + 1 = 3.$

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Step 4: Combine the results

Now, sum all the terms: $3 - 3 + 3 = 3.$

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Final Answer:

$\boxed{3}$

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Do you have questions or need further clarification?

Related Questions:

1. What are the steps for integrating polynomials?
2. How does substitution affect definite integrals?
3. Can definite integrals ever yield negative results?
4. How do you verify the result of an integral graphically?
5. What is the significance of definite integrals in real-world applications?

Tip:

Always double-check the limits of integration and signs when solving definite integrals!