Evaluate \(\int^{1}_{-2}(x-1)^{2}dx\)

To evaluate the integral

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

We first expand the integrand:

\((x - 1)^{2} = x^{2} - 2x + 1\)

Thus, the integral becomes:

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

Now, we evaluate each integral:

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

\(\left[ \frac{x^{3}}{3} \right]_{-2}^{1} = \frac{1}{3} - \left(-\frac{8}{3}\right) = 3\)

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

\(-2\left[ \frac{x^{2}}{2} \right]_{-2}^{1} = -2\left(\frac{1}{2} - 2\right) = 3\)

For \(\int_{-2}^{1} 1 \, dx\):

\([x]_{-2}^{1} = 1 + 2 = 3\)

Combining the results: 3 + 3 + 3 = 9.

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