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Consider the region $R$ which is the region bounded between the four points $$ (0,b) \ , \ \ (a,a+b)\ , \ \ (a,c) \ , \ \ (0,c) $$ where $0<a<b$ and $a+b<c$, which looks like:

enter image description here

For a function $f(x,y)$ I would like to write the integral $$ \iint_R f(x,y) dx dy $$ as a single integral.

I can write this as a sum of two terms (over a triangular and rectangular region): $$ \iint_R f(x,y) dx dy = \overbrace{\int_0^a dx \int_{x+b}^{a+b} dy\ f(x,y) }^{\text{triangle between }(0,b)\ \& \ (0,a+b) \ \& \ (a,a+b) }\ \ + \ \ \overbrace{\int_0^a dx \int_{a+b}^{c} dy\ f(x,y)}^{\text{rectangle between }(0,a+b)\ \& \ (a,a+b) \ \& \ (a,c) \ \& \ (0,c) } $$

QUESTION: Is there a way to write this integral without splitting it apart into different regions?

I am tempted to write the above as $$ \iint_R f(x,y) dx dy = \int_0^a dx \int_{x+b}^{c} dy\ f(x,y) $$ but I don't think this is correct.

QuantumEyedea
  • 2,805

2 Answers2

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What you are tempted to write is correct. Please note the region is bound between vertical lines $x = 0$ and $x = a$ and horizontal lines $y = c ~ $ and $ ~ y = x + b$.
(Based on your diagram, assuming $c \gt a + b \gt b$).

So the double integral can indeed be set up as,

$ \displaystyle \int_0^a\int_{x+b}^c f(x, y) ~ dy ~ dx$

Math Lover
  • 51,819
2

Yes, what you have is correct. Recall some integral rules:

\begin{align*} \int_a^b f(x)\ dx + \int_a^b g(x)\ dx&=\int_a^b [f(x)+ g(x)]\ dx\\ \int_a^bf(x)\ dx +\int_b^c f(x)\ dx&=\int_a^c f(x)\ dx \end{align*}

Notice that your first result is exactly your second result:

\begin{align*} \int_0^a \int_{x+b}^{a+b} f(x,y)\ dy\ dx+\int_0^a \int_{a+b}^{c} f(x,y)\ dy\ dx &=\int_0^a\left[ \int_{x+b}^{a+b} f(x,y)\ dy+ \int_{a+b}^{c} f(x,y)\ dy\right]\ dx\\ &=\int_0^a \int_{x+b}^{c} f(x,y)\ dy\ dx \end{align*}

Bonnaduck
  • 4,058