Solving a Double Integral over a Rectangle with Given Vertices

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SUMMARY

The discussion centers on evaluating the double integral \(\int\ \int_R\ x^2e^ydA\) over the rectangle R defined by the vertices (0,0), (1,0), (1,3), and (0,3). The correct evaluation yields \(\frac{1}{3}(e^3 - 1)\), confirming the simplicity of the integral due to the rectangular area. Participants agree that the problem's straightforward nature is typical in assessments to gauge fundamental understanding of double integrals. The overall sentiment is supportive, encouraging confidence in solving such problems.

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  • Understanding of double integrals
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  • Knowledge of exponential functions
  • Basic geometry of rectangles in coordinate systems
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Gwilim
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Evaluate \int\ \int_R\ x^2e^ydA

Over the rectangle R with vertices (0,0), (1,0), (1,3) and (0,3).

My answer:

\int\ \int_R\ x^2e^ydA = \int_0^3\ \int_0^1\ x^2e^ydA
= \int_0^3\ [x^3/3]_0^1 e^y dy
= 1/3 \int_0^3\ e^ydy
= 1/3 (e^3-1)

Double integrals are new to me, so if someome could check my answer that would be greatly helpful
 
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looks ok.
 
seems too easy for 10 marks. There's barely three lines of working there.
 
Gwilim said:
seems too easy for 10 marks. There's barely three lines of working there.

well I don't know if 10marks is much or not. The integral is easy to solve since you had a rectangular area.
 
malawi_glenn said:
well I don't know if 10marks is much or not. The integral is easy to solve since you had a rectangular area.

The whole 2 hour paper has 100 marks in total. Anyway, thanks for the confirmation.
 
You are definitely correct. As for the facility with which you did this problem, you're just a superstar at this stuff ;)

Sometimes profs will toss in easy questions to discern who has, at least, a basic command of the principles involved from those who don't even know what an integrand is.
 
Have confidence! GJ :)
 

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