Pulling out a constant in double integrals $$ \iint {3x-y\over 9} \mathrm{d}x\mathrm{d}y$$
Is it safe to pull out a constant such as:
$$ {1\over 9}\iint (3x-y) \ \mathrm{d}x\mathrm{d}y$$
I know this sounds silly, and it should be obvious that you can do this. But when I was trying to solve this integral for $0 \lt x \lt 2$ and $0 \lt y \lt 1$:
$$ \iint \left(\frac{y(1+3y^2)}{4}\right) \mathrm{d}x\mathrm{d}y$$
The answer to the above integral should be $\frac58$, but my initial answer was $\frac54$ even after multiplying the $\frac14$ constant. But after multiplying $\frac14$ constant again, I got $\frac58$ as my answer.
 A: Yes. The constant multiple comes out easily. Just as it would for a sum of finite terms. Check your calculus!
A: The integral is a linear operator. This means you can always do this, even if you apply two linear operators in a row.
A: $$\begin{align}\int_{0}^{1}\int_{0}^{2} \left(\frac{y(1+3y^2)}{4}\right) \mathrm{d}x\;\mathrm{d}y
 & = \tfrac 1 4\cdot\int_0^2  \mathrm{d}x\cdot\int_0^1 (y+3y^3)\;\mathrm{d}y
\\ & = \tfrac 1 4\cdot 2\cdot{\big[\tfrac 1 2 y^2+\tfrac 3 4 y^4\big]}_{y=0}^{y=1} 
\\ & = \tfrac 1 2\cdot\tfrac 5 4
\\ & = \tfrac 5 8
\end{align}$$
A: HINT:
$$\int\int\frac{3x-y}{9}\text{d}x\text{d}y=$$
$$\int\left(\int\frac{3x-y}{9}\text{d}x\right)\text{d}y=$$
$$\int\left(\frac{1}{9}\int\left(3x-y\right)\text{d}x\right)\text{d}y=$$
$$\int\left(\frac{1}{9}\left(3\int x\text{d}x-y\int 1\text{d}x\right)\text{d}x\right)\text{d}y=$$
$$\int\left(\frac{1}{9}\left(3\cdot\frac{1}{2}x^2-y\cdot x+C_1\right)\text{d}x\right)\text{d}y=$$
$$\int\left(\frac{1}{9}\left(\frac{3x^2}{2}-xy\right)+C_1\right)\text{d}y=$$
$$\frac{1}{9}\int\left(\frac{3x^2}{2}-xy+C_1\right)\text{d}y=$$
$$\frac{1}{9}\left(\frac{3x^2}{2}\int 1\text{d}y-x\int y\text{d}y+C_1\int 1\text{d}y\right)=$$
$$\frac{1}{9}\left(\frac{3yx^2}{2}-\frac{xy^2}{2}+yC_1+C_2\right)=$$
$$\frac{1}{9}\left(\frac{3yx^2}{2}-\frac{xy^2}{2}\right)+yC_1+C_2$$
