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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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    Thank you Zev, much appreciated.2011-04-22

1 Answers 1

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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 $0$.

The method for the second integral looks good.

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., $\int_0^5(x+1)^2dx=\frac{1}{3}(5+1)^3-\frac{1}{3}(0+1)^3=\frac{216}{3}-\frac{1}{3}$.

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    @Ocasta: Glad to help. I suspected it was a habit picked up due to frequent evaluations like $\int_0^b x^n=\frac{1}{n+1}b^{n+1}$, $n\gt -1$.2011-04-22