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I want to change the order of integration of the following

$$\int_{1}^{+ \infty} \left (\int_{1}^{\sqrt{y}} x^3e^{-xy} dx \right) dy.$$

I get the bounds $1 \le y \le x^2$ and $1 \le x \le \infty$, and so the integration becomes

$$\int_{1}^{+ \infty} \left (\int_{1}^{x^2} x^3e^{-xy} dy \right) dx.$$

As shown below, the area which is integrated is $I$, bounded by the curve $x=\sqrt{y}$, $x=1$ and $y=1$. Is this correct? Also, in general, how do I know which area which is integrated? Sometimes it is clear which area is integrated, sometimes not. Thank you for your time.

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user4167
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    Good picture! It tells you everything. The first integral is over region I. If we first integrate with respect to $y$, then $y$ will travel from the curve, that is, $x^2$, to $\infty$. – André Nicolas Jun 01 '13 at 20:32

2 Answers2

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You absolutely have to figure out which region you're doing to get the limits right. Draw (by hand) a horizontal line with $y$ fixed, and shade the $x$ values that go from $1$ to $\sqrt y$. What is your region? Then change the order and do it in reverse: what are the possible $x$ values in the region, and for a fixed $x$, what does $y$ do [i.e., draw a vertical line segment and figure out its endpoints]?

Ted Shifrin
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I think you should get $+\infty > y\geq x^{2}$ so the inner integral has different boundaries.

Neph
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