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Let's say I have the following integral:

$$\int_0^1\int_1^3 x+y\: dy\,dx $$

Can I separate this (like with multiplication, i.e x*y) to:

$$\int_0^1 x\:dx +\int_1^3 y\: dy $$

Is this correct?

tripleee
  • 135
Dakalaom
  • 121
  • No, you can't do that. But you can use the linearity of the integral to separate it as a sum of two double integrals which are very easy to calculate. – Laz Mar 06 '22 at 19:15

3 Answers3

5

No, it is not.

The property which you may be interested in is given by \begin{align*} \int_{a}^{b}\int_{c}^{d}f(x)g(y)\mathrm{d}y\mathrm{d}x = \int_{a}^{b}f(x)\mathrm{d}x\int_{c}^{d}g(y)\mathrm{d}y \end{align*}

At the given example, one has that \begin{align*} \int_{0}^{1}\int_{1}^{3}(x + y)\mathrm{d}y\mathrm{d}x & = \int_{0}^{1}\left(xy + \frac{y^{2}}{2}\right)\Biggr|_{y=1}^{y=3}\mathrm{d}x\\\\ & = \int_{0}^{1}(2x + 4)\mathrm{d}x\\\\ & = \left(x^{2} + 4x\right)\Biggr|_{0}^{1} = 5 \end{align*}

On the other hand, the other integrals result into the following value: \begin{align*} \int_{0}^{1}x\mathrm{d}x + \int_{1}^{3}y\mathrm{d}y = \frac{1}{2} + 4 = \frac{9}{2} \end{align*}

Hopefully this helps !

Átila Correia
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4

No.

$$\int_0^1\int_1^3 (x+y) dydx = \int_0^1\int_1^3 x dydx + \int_0^1\int_1^3 ydydx = \int_1^3\int_0^1 x dxdy + \int_0^1\int_1^3 ydydx$$

is how you use linearity for this integral.

Doge Chan
  • 871
0

In general \begin{eqnarray*}\int_a^b \int_c^d f(x)+g(y) \quad dy\, dx&=& \int_a^b \int_c^d f(x) \quad dy\, dx+\int_a^b \int_c^dg(y) \quad dy\, dx\\\\ &=&(d-c)\int_a^b f(x) \quad dx +(b-a)\int_c^d g(y) \quad dy. \end{eqnarray*}

Thus in your case: $$\int_0^1\int_1^3 x+y\: dy\,dx =2\int_0^1 x\:dx +\int_1^3 y\: dy,$$ which is almost exactly what you conjectured.

tkf
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