Evaluate Double integral over triangular region

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SUMMARY

The discussion focuses on evaluating the double integral of the function \( ye^x \) over a triangular region defined by the vertices (0,0), (2,4), and (6,0). The correct limits of integration are established as \( 0 \) to \( 4 \) for \( y \) and \( 0.5y \) to \( 6-y \) for \( x \). The integral simplifies to \( \int_{0}^{4} (ye^{6 - y} - ye^{y/2}) dy \). The recommended method for solving this integral is to apply integration by parts on both terms.

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  • Knowledge of exponential functions and their properties
  • Ability to interpret triangular regions in coordinate geometry
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  • Study the application of integration by parts in different contexts
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nlsherrill
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Homework Statement


Evaluate the double integral.

[I don't know how to write it in latex, sorry, but its the double integral over the region 'd' of ye^x dA

D is the triangular region with vertices (0,0), (2,4), and (6,0).


Homework Equations





The Attempt at a Solution



So the limits of integration I have are .5y to 6-y with respect to x, and 0 to 4 with respect to y. I think this is right. The farthest I can get is:

Integral 0..4 ye^6-y - ye^.5y dy

I can't figure out how to integrate these terms. Any hints?
 
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nlsherrill said:

Homework Statement


Evaluate the double integral.

[I don't know how to write it in latex, sorry, but its the double integral over the region 'd' of ye^x dA

D is the triangular region with vertices (0,0), (2,4), and (6,0).


Homework Equations





The Attempt at a Solution



So the limits of integration I have are .5y to 6-y with respect to x, and 0 to 4 with respect to y. I think this is right. The farthest I can get is:

Integral 0..4 ye^6-y - ye^.5y dy

I can't figure out how to integrate these terms. Any hints?
Here's your integral in LaTeX:
[tex]\int_{y = 0}^4 (ye^{6 - y} - ye^{y/2})dy[/tex]

Break this up into two integrals. Note that e6 - y = e6e-y

Integration by parts should work on both integrals.
 
Thanks! forgot about integration by parts.
 

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