Although you probably will not need to do this on the actuarial exam, it is very good practice to change an integral set up \(dx\ dy\) to an integral set up \(dy \ dx\) or vice versa. Consider the integral
\(\int_0^1 \int_y^{\sqrt y} f(x,y)\ dx\ dy\)
Can we write an equivalent integral in the form \(\int_?^? \int_?^{?} f(x,y)\ dy\ dx\)? It really helps to sketch the region. The lower limit on \(x\) is the curve \(x=y\), and the upper limit on \(x\) is the curve \(x=\sqrt y\), which can be written as \(y=x\) and \(y=x^2\) respectively when \(y\) is in the interval \([0,1]\). Therefore if we reverse the order of integration, we obtain the integral
\(\int_0^1 \int_{x^2}^x f(x,y)\ dy \ dx.\)
Now you try one. Reverse the order of integration in the integral
\(\int_{1}^2 \int_4^{4x} f(x,y)\ dy \ dx.\)
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\(\int_{1}^2 \int_4^{4x} f(x,y)\ dy \ dx = \int_{4}^8 \int_{y/4}^{2} f(x,y)\ dx \ dy\)