# Difficult double integration

I am trying to integrate: $$\int \int_D (3x+4y)^4 dx dy$$ $$0 \leq x \leq 4$$$$0 \leq y \leq 1$$ I see it looks quite easy but I can't manage to integrate it.

1. I tried to expand the polynom and integrate it for $dx$, and then for $dy$. It grew obviously very fast out of control:

$$(3x)^4+4(3x)^3(4y)+6(3x)^2(4y)^2+4(3x)(4y)^3+(4y)^4$$, that to be integrated wrt to $x$, and then to $y$, seems wrong.

1. I tried to rewrite that as: $$\int \int_D (3rcos \theta +4r sin \theta)^4 dr d\theta$$ $$\int \int_D r^4(3cos \theta +4sin \theta)^4 dr d\theta$$

I then thought that I need to integrate wrt $r$, with $r$ going from zero to 5:

$$\int \int_D 5e^{i\theta} dr d\theta$$

1. I tried to replace $(3x+4y)^4$ with $u^4$, but when I integrate from 0 to 1 (wrt $y$), I get $\dfrac{1}{5}x$, so I guess it's wrong as well.

But then I don't know the angle I am supposed to integrate.

I have no further ideas about how to proceed with this problem.

• What is $D$ here ? Jun 28, 2018 at 8:40
• Please see my edit! Jun 28, 2018 at 8:43
• What’s so out of control in your first try? There are only 5 terms and each is essentially a constant times a power of the variable with respect to which you’re integrating. There’s no need to expand, though. The antiderivatives can be computed directly.
– amd
Jun 28, 2018 at 8:44
• Then, with such a $D$, the first try is by far the easiest one. Jun 28, 2018 at 8:45
• hint: $\int (3x+4y)^4 dx=\frac{(3x+4y)^5}{15}+C$ Jun 28, 2018 at 8:47

We solve explicitly, using Fubini: \begin{aligned} \int \int_D (3x+4y)^4 \;dx\; dy &= \int_0^4dx \int_0^1 (3x+4y)^4\; dy \\ &= \int_0^4dx \left[\ \frac 15\cdot\frac 14(3x+4y)^5\ \right]_0^1\; dy \\ &= \int_0^4 \frac 15\cdot\frac 14\Big[\ (3x+4)^5-(3x+0)^5\ \Big]\; dx \\ &= \frac 15\cdot\frac 14\cdot \frac 16\cdot\frac 13\Big[\ (3x+4)^6-(3x+0)^6\ \Big]_0^4\; dx \\ &= \frac 15\cdot\frac 14\cdot \frac 16\cdot\frac 13\Big[\ 18^6-4^6-12^6+0^6\ \Big] \ . \end{aligned} Computer check, here using sage:

sage: var( 'x,y' );
sage: integral( integral( (3*x+4*y)^4, y, 0, 1), x, 0, 4 )
191488/5
sage: ( (12+4)^6 - (12+0)^6 - (0+4)^6 + (0+0)^6 ) / ( 5*4*6*3 )
191488/5

• I'm sorry, what did you do there? Did you just not developed Pascal triangle? Or did you do it, but not on screen? Jun 28, 2018 at 12:13
• There is no Pascal triangle involved, i cannot see any connection. I have just integrated in the most simple way one can do it. The primitive of the function $x\to (ax+b)^n$ is $$\frac 1{n+1}\cdot\frac 1a(ax+b)^{n+1}\ ,$$ so i applied this twice, once while integrating w.r.t. $y$, then again while integrating w.r.t. $x$. The "primitive of the primitive" has then to be evaluated "in the four corners" of the given solid rectangle, and we are done. Jun 28, 2018 at 16:23
• I never saw that Fubini method before! Does it work all the time? Jun 29, 2018 at 4:01
• For a rectangular domain of integration, when the integral with respect to one of the variables is "simple", and the result of this integration is an easy game for the integration w.r.t. the second variable, yes, always use Fubini! Jun 29, 2018 at 16:55
• I think that both your reply and @Rumplestillskin were brilliant. I vote yours as answer because we did not learn Fubini at school and I thought it was amazing! Jun 30, 2018 at 4:53

HINT

$$\int (3x+4y)^4dx = \frac{(3x+4y)^5}{15} + C.$$

Hence, you can integrate this equation as it is

$$\int_0^1 \int_0^4 (3x+4y)^4 dxdy,$$

becomes

$$\frac{1024}{5}\int_0^1 \left[ 5y^4 + 30y^3 + 90y^2 + 135y + 81 \right] dy.$$

• Thank you, it was NOT that difficult once you all explained. Can you please give me ideas on alternative solving techniques? Jun 28, 2018 at 12:12
• @Dovendyr Why would you want to use some other method when this direct method gets you the solution in two simple steps?
– amd
Jun 28, 2018 at 17:24
• Because I have even more difficult integrals coming, so I want to see some other ways to go. Are you sure there is nothing to do with that $3x+4y$? It seems like a classical circle with radius 5? Jun 29, 2018 at 4:03
• Give an example of some more difficult integrals and I’m sure someone will be able to help with what you don’t understand. Jun 29, 2018 at 4:18
• I'll be back with some more, I'm note sure I can put them in the comment section! Jul 2, 2018 at 9:10