Question

In evaluating a double integral over a region\(D\), sum of iterated integrals was obtained as follows:

\(\int {\int\limits_D {f\left( {x,y} \right)} } dA = \int\limits_0^1 {\int\limits_0^{2y} {f\left( {x,y} \right)} } dxdy + \int\limits_1^3 {\int\limits_0^{2 - y} {f\left( {x,y} \right)dxdy} } \).

Sketch the region\(D\)and express the double integral as an iterated integral with reversed order of integration.

Short answer

The process of finding a function\(g\left( x \right)\) the derivate of which, \(Dg\left( x \right)\)is equal to the given function\(f\left( x \right)\).

Step-by-step solution

Given data

For the first integral the limit of integration Indicate that\(0 \le x \le 2y\)and then set of inequalities it valid for\(0 \le y \le 1\).

Graph:

Double integral:

The union of the two original regions of integration is a type\(1\)region and the two integrals.

Note that a vertical line pasting the line\(y = \frac{x}{2}\)and leaves at the line\(y = 3 - x\).

\(\begin{aligned}\int {\int\limits_D {f\left( {x,y} \right)} } dA &= \int\limits_0^1 {\int\limits_0^{2y} {f\left( {x,y} \right)} } dxdy + \int\limits_1^3 {\int\limits_0^{2 - y} {f\left( {x,y} \right)dxdy} } \\ &= \int\limits_0^2 {\int\limits_{{x \mathord{\left/ {\vphantom {x 2}} \right.} 2}}^{3 - x} {f\left( {x,y} \right)} } dxdy\end{aligned}\)

Therefore, the integral is \(\int\limits_0^2 {\int\limits_{{x \mathord{\left/ {\vphantom {x 2}} \right.} 2}}^{3 - x} {f\left( {x,y} \right)} } dxdy\).