\( \int _{}^{}\int _{-1}^{2}(x-2) d x d \)
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To evaluate the double integral of \( \int_{-1}^{2}(x-2) dx \), we first need to find the inner integral \( \int(x-2) dx \) over the range from -1 to 2. Calculating this gives: \[ \int(x-2) dx = \frac{x^2}{2} - 2x + C \] Now we evaluate it between the limits -1 and 2: \[ \left[ \frac{(2)^2}{2} - 2(2) \right] - \left[ \frac{(-1)^2}{2} - 2(-1) \right] = \left[ 2 - 4 \right] - \left[ \frac{1}{2} + 2 \right] = -2 - \left[ \frac{1}{2} + 2 \right] = -2 - 2.5 = -4.5 \] Thus, the value of the integral is -4.5.