# How to check in Wolfram|Alpha double integral?

Posted 20 days ago
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| D: x=y^2 and x-y=2 Answer
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Posted 20 days ago
 Plot[{x^2, x + 2}, {x, 0, 3}] Integrate[x + 2 - x^2, {x, 0, 2}] Answer
Posted 20 days ago
 it's wrong, another shape is bounded by these lines. y^2=x and x-y=2  Answer
Posted 20 days ago
 1) you didn't specify your problem exactly2) I interchanged x and y, So your figure is basically the same as my region3) Try Plot[{x^2, x + 2}, {x, -2, 3}] Integrate[x + 2 - x^2, {x, -1, 2}] Answer
Posted 20 days ago
 Does this match what you expect? Integrate[Integrate[y^2,{x,xlow(y),xhigh(y)}],{y,ylow,yhigh}] so Integrate[Integrate[y^2,{x,y^2,2+y}],{y,-1,2}] Giving exactly that to either WolframAlpha or to Mathematica returns 63/20==3.15It is also possible to do the same with Integrate[y^2,{y,-1,2},{x,y^2,2+y}] but the y and x must be given in exactly that order if you use that form.As a check, If you are using Mathematica or other calculation tools then it is possible to calculate a crude manual approximation of this total=0;d=1/100; For[y=-1,y<=2,y=y+d, For[x=y^2,x<=2+y,x=x+d, total=total+y^2*d^2 ]]; total//N which returns 3.1652 and that is approximately the same as the result from the integration and smaller values for d will give better approximations..But after doing that be sure to tell Mathematica to forget the cached assigned values to x and y with x=.;y=.; before doing further Integrate which do not expect to have x or y have been assigned values. Answer
Posted 19 days ago
 Thx you for your answers!! Answer
Posted 19 days ago
 Here is another approach Integrate[Integrate[1, {x, y^2, y + 2}], {y, -1, 2}] Or did you mean this? Integrate[Integrate[y^2, {x, y^2, y + 2}], {y, -1, 2}] Answer
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