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$$ \iint {3x-y\over 9} \mathrm{d}x\mathrm{d}y$$

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.

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  • $\begingroup$ Put your limits in please so we can see how you got $\frac58$? $\endgroup$
    – BLAZE
    Commented Oct 16, 2015 at 12:09

4 Answers 4

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Yes. The constant multiple comes out easily. Just as it would for a sum of finite terms. Check your calculus!

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  • $\begingroup$ Then why is it that my second double integral problem has to multiply by the constant (1/4) twice? $\endgroup$
    – dendritic
    Commented Oct 16, 2015 at 12:03
  • $\begingroup$ ? I am not sure what you mean by this. As you have written your double integral, I see only one factor of $(1/4)$. Further your integral has no $x$ in the integrand so, really is just a single integral. $\endgroup$
    – amcalde
    Commented Oct 16, 2015 at 12:05
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    $\begingroup$ The answer is indeed $5/8$ but I think you've just made an error. $\endgroup$
    – amcalde
    Commented Oct 16, 2015 at 12:08
  • $\begingroup$ Well, I'm an idiot. I just found my problem. I added my fractions incorrectly at the very end. $\endgroup$
    – dendritic
    Commented Oct 16, 2015 at 12:08
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    $\begingroup$ Happens to the best of us. No worries. $\endgroup$
    – amcalde
    Commented Oct 16, 2015 at 12:09
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The integral is a linear operator. This means you can always do this, even if you apply two linear operators in a row.

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$$\begin{align}\int_{0}^{1}\int_{0}^{2} \left(\frac{y(1+3y^2)}{4}\right) \mathrm{d}x\;\mathrm{d}y & = \tfrac 1 4\cdot\int_0^2 \mathrm{d}x\cdot\int_0^1 (y+3y^3)\;\mathrm{d}y \\ & = \tfrac 1 4\cdot 2\cdot{\big[\tfrac 1 2 y^2+\tfrac 3 4 y^4\big]}_{y=0}^{y=1} \\ & = \tfrac 1 2\cdot\tfrac 5 4 \\ & = \tfrac 5 8 \end{align}$$

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HINT:

$$\int\int\frac{3x-y}{9}\text{d}x\text{d}y=$$ $$\int\left(\int\frac{3x-y}{9}\text{d}x\right)\text{d}y=$$ $$\int\left(\frac{1}{9}\int\left(3x-y\right)\text{d}x\right)\text{d}y=$$ $$\int\left(\frac{1}{9}\left(3\int x\text{d}x-y\int 1\text{d}x\right)\text{d}x\right)\text{d}y=$$ $$\int\left(\frac{1}{9}\left(3\cdot\frac{1}{2}x^2-y\cdot x+C_1\right)\text{d}x\right)\text{d}y=$$ $$\int\left(\frac{1}{9}\left(\frac{3x^2}{2}-xy\right)+C_1\right)\text{d}y=$$ $$\frac{1}{9}\int\left(\frac{3x^2}{2}-xy+C_1\right)\text{d}y=$$ $$\frac{1}{9}\left(\frac{3x^2}{2}\int 1\text{d}y-x\int y\text{d}y+C_1\int 1\text{d}y\right)=$$ $$\frac{1}{9}\left(\frac{3yx^2}{2}-\frac{xy^2}{2}+yC_1+C_2\right)=$$ $$\frac{1}{9}\left(\frac{3yx^2}{2}-\frac{xy^2}{2}\right)+yC_1+C_2$$

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