Question
Evaluate the integrals.$$\int_{0}^{1} \int_{0}^{3-3 x} \int_{0}^{3-3 x-y} d z d y d x$$
Step 1
The limits of $z$ are from 0 to $3-3x-y$. So, we have $$\int_{0}^{1} \int_{0}^{3-3 x} \int_{0}^{3-3 x-y} d z d y d x = \int_{0}^{1} \int_{0}^{3-3 x} (3-3x-y) d y d x.$$ Show more…
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