Math 2.0

Real functions Concept

Integration over a non-rectangular region

Builds on Double integralIterated integralJordan measure

Summary

To integrate a function ff over a bounded region DD that is not a rectangle, take a rectangle R⊇DR \supseteq D. Then integrate over RR the function equal to ff on DD and to 00 on R∖DR \setminus D. For a region between two graphs, D={(x,y):a≤x≤b, φ1(x)≤y≤φ2(x)}D = \{(x, y) : a \le x \le b, \ \varphi_1(x) \le y \le \varphi_2(x)\} with φ1≤φ2\varphi_1 \le \varphi_2 continuous, a continuous ff gives ∬Df dA=∫ab∫φ1(x)φ2(x)f(x,y) dy dx\iint_D f \, dA = \int_a^b \int_{\varphi_1(x)}^{\varphi_2(x)} f(x, y) \, dy \, dx.

Remarks

Such a region is normal with respect to the xx-axis. A region of the form {c≤y≤d, ψ1(y)≤x≤ψ2(y)}\{c \le y \le d, \ \psi_1(y) \le x \le \psi_2(y)\} is normal with respect to the yy-axis. The definition does not depend on the choice of RR. For continuous ff, the integral exists when the boundary of DD has Jordan content zero. Example: for D={0≤x≤1, x2≤y≤1}D = \{0 \le x \le 1, \ x^2 \le y \le 1\}, ∬Dx dA=∫01x(1−x2) dx=1/4\iint_D x \, dA = \int_0^1 x (1 - x^2) \, dx = 1/4. The double integral over DD is computed as an iterated integral with variable bounds.

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  1. Euclidean space object
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Further reading

  1. Web https://en.wikipedia.org/wiki/Multiple_integral, Double integral over a normal domain
  2. Web https://en.wikipedia.org/wiki/Multiple_integral, Mathematical definition
About this page
Tag0TXCQ9CST1
URIm2://0TXCQ9CST1
Aliasesintegral-over-a-normal-domain, type-i-and-type-ii-regions
Kindconcept
WritingW0 stub
MSC26B15
UsesDouble integral, Iterated integral, Jordan measure

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