As the title asks. the boundary is xmin=y/4 and xmax=6-0.5*y, ymin=0, and ymax=4

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Roger Stafford
Roger Stafford am 11 Apr. 2016
Bearbeitet: Roger Stafford am 11 Apr. 2016
This is a problem you can work by hand using ordinary methods of integral calculus to get a precise answer. Probably the symbolic toolbox can do the same.

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jgg
jgg am 11 Apr. 2016

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I think this should work; you want the integral2 function.
fun = @(x,y)(x.*exp(y));
ymin = @(x)(x/4);
ymax = @(x)(6 - 0.5*x);
q = integral2(fun,0,4,ymin,ymax)

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