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.