1. Evaluate (𝑥 − 3) 𝑑 where = , , − 1 ≤ ≤ 1, 0 ≤ ≤ 2, 0 ≤ ≤ 1 }.
∫z∫y∫x(x−3)dxdydz∫01∫02∫−11(x−3)dxdydz∫01∫02x2−3x∣−11dydz∫01∫02−6dydz∫01(−6y∣02)dz∫01−12dz=−12z∣01=−12\int_z\int_y\int_x(x-3)dxdydz\\\int_0^1\int_0^2\int_{-1}^1(x-3)dxdydz\\ \int_0^1\int_0^2x^2-3x|_{-1}^1dydz\\\int_0^1\int_0^2-6dydz\\\int_0^1(-6y|_0^2)dz\\\int_0^1-12dz=-12z|_0^1=-12\\∫z∫y∫x(x−3)dxdydz∫01∫02∫−11(x−3)dxdydz∫01∫02x2−3x∣−11dydz∫01∫02−6dydz∫01(−6y∣02)dz∫01−12dz=−12z∣01=−12
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