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$$I:=\int_{0}^{1} \int_{x}^{1}y^4e^{xy^2}dy dx$$ Here the region of integration is the triangle with vertices $(0,0),(0,1)$ and $(1,1)$ and given as a type-1 region. We can convert it into a type-2 region which makes the integral easier. $$I=\int_{0}^{1}\int_{0}^{y}y^4e^{xy^2}dx dy=\int_{0}^{1} y^2(e^{y^3}-1)dy=\frac {e-2}{3}$$ Is this correct? I'd like to add graphs but I'm still learning how to do that.

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    $\begingroup$ Yes, you have it. Well done. $\endgroup$ – Mark Viola Jan 30 at 4:37
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Basically , the integral means you have to sum the differential term (function ) for all the points in the given area . Now in the 1st equation , the integral first summed it for all points with common x coordinate (for them , x becomes a constant) , and the value y takes is from x to 1( x is a constant). This creates sums for the strips in terms of x coordinate of the strip. Then when we integrate again , these strips are summed up over all x.

The second equation ,finds the sum in strips with common y coordinate (x goes from 0 to y in a strip). And then , similarly , strips are summed up for all y . Both of them are equivalent and yes, the second one an easy one to evaluate.

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Here's a graph, for your better understanding:

enter image description here

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