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I´m not sure how to integrate this, should I do a change of variables (we haven't seen it yet in our course)? Is there any other way?

$$ \int_0^2\int_0^1 (x+y)^8 dx dy $$

I'm rather interested in the general case, but this is the integral that motivated the question.

EDIT: I know expanding this using the binomial theorem works. However I don't think this is the intent of the exercise. As I said, I want to know about some technique to use in more general cases.

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  • $\begingroup$ Consider y to be constant in the inner integral and work your way outward. $\endgroup$
    – user328442
    Commented Nov 9, 2016 at 19:58
  • $\begingroup$ For this particular integral, the easiest way is to expand $(x+y)^8$, and then you have a sum of 9 products of two simple 1D integrals $\int_0^2x^8dx\int_0^1dy+8\int_0^2x^7dx\int_0^1ydy+...$ $\endgroup$
    – Andrei
    Commented Nov 9, 2016 at 19:59
  • $\begingroup$ I didn't want to do that. Also, I think this is supposed to invite us to use some technique. $\endgroup$
    – GoetheGrimm
    Commented Nov 9, 2016 at 20:00
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    $\begingroup$ I do not understand the scope of this question. You do now want to perform a binomial expansion, since it involves a lot of terms. You do not want to perform a change of variables, since you haven't seen yet, so what are you likely to accept from us? $\endgroup$
    – Jack D'Aurizio
    Commented Nov 9, 2016 at 20:03
  • $\begingroup$ Well, I don't know, which is why I ask. If a change of variable is the standard approach, then I suppose it answers my question. Again, I don't know of another way, am asking whether there's one. $\endgroup$
    – GoetheGrimm
    Commented Nov 9, 2016 at 20:12

3 Answers 3

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$$I=\int_{0}^{2}\int_{0}^{1}(x+y)^8\,dx\,dy = \int_{0}^{2}\left. \frac{(x+y)^9}{9}\right|_{x=0}^{x=1}\,dy =\frac{1}{9}\int_{0}^{2}\left[(y+1)^9-y^9\right]\,dy$$ hence $I= \color{red}{\large\frac{3^{10}-2^{10}-1}{90}}$.

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We can evaluate the first integral $$ \int_0^1(x+y)^8dx $$ with the substitution $x+y=t \quad \rightarrow \quad dx=dt$ so, the integral becomes: $$ \int_{y}^{1+y}t^8dt=\frac{1}{9}\left[ (1+y)^9-y^9\right] $$ and the second integration becomes: $$ \frac{1}{9}\left[\int_0^2 (1+y)^9dy-\int_0^2y^9dy\right] $$

that is not difficult to solve using an analogous substitution for the first integral.

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Hint

$$I=\sum_{k=0}^8\binom{8}{k}\int_0^2x^kdx \int_0^1y^{8-k}dy$$

$$=\sum_{k=0}^8\binom{8}{k}\frac{2^{k+1}}{k+1}\frac{1}{9-k}$$

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