# Problem with Repeated Integrals

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.

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I've put your equations in $\LaTeX$; hopefully I have preserved their content correctly. –  Zev Chonoles Apr 22 '11 at 21:16
Thank you Zev, much appreciated. –  Ocasta Eshu Apr 22 '11 at 23:23
The first integral is incorrect because you evaluated the antiderivative only at $x+y$. You either forgot to evaluate at $0$ or incorrectly found that evaluation to be $0$.
For the third integral you made the same mistake as for the first. Evaluation at $0$ does not mean that the value is zero. $\int_0^bF'(x)dx=F(b)-F(0)$, which is not $F(b)$ unless $F(0)=0$. E.g., $\int_0^5(x+1)^2dx=\frac{1}{3}(5+1)^3-\frac{1}{3}(0+1)^3=\frac{216}{3}-\frac{1}{3}$.
@Ocasta: Glad to help. I suspected it was a habit picked up due to frequent evaluations like $\int_0^b x^n=\frac{1}{n+1}b^{n+1}$, $n\gt -1$. –  Jonas Meyer Apr 22 '11 at 23:43