I havent had the time to familiarize myself with Latex quite yet, so please excuse my formatting. I have attempted the following problem four times and got four completely different answers.
$\int_0^1\int_1^2\int_0^{x+y}12(4x+y+3z)^2 dz dy dx$
to my understanding, the first integral should equal:
$\frac{4}{3}(7x+4y)^3$
The second would be:
$\frac{1}{12}(7x+8)^4-\frac{1}{12}(7x+4)^4$
And the final integral:
$\frac{1}{420}(7+8)^5-\frac{1}{420}(7+4)^5$
or 1424.59
Again I've tried several different methods receiving different answers, each marked as wrong on the homework website. I think I'm missing something basic here, but I dont know what.