## Multivariate Calculus for Actuarial Science, Part II

### Setting Up an Integral

Of course, you usually have to set up the interated integral yourself. For example you may be asked to evaluate the integral

\(\iint_R \left(x+y\right) \, dA,\)

where \(R\) is the rectangle bounded by \(x=1\), \(x=2\), \(y=0\), and \(y=2\). In this case, it is clear that the variable \(x\) is varying from \(1\) to \(2\), and the variable \(y\) is varying from \(0\) to \(2\). You can think of determining the bounds on \(x\) by drawing a horizontal line through the rectangle \(R\), and determining the bounds on \(y\) by drawing a vertical line through the rectangle \(R\). It really doesn't matter whether you decide to integrate \(dx \ dy\) or \(dy \ dx\). We need to evaluate one of these integrals:

\(\iint_R \left(x+y\right) \, dA=\int_1^2 \int_0^2 \left(x+y\right)dy\ dx = \int_0^2 \int_1^2 \left(x+y\right)dx\ dy.\)

Let us integrate \(dx\) first.

\(\int_0^2 \int_1^2 \left(x+y\right)dx\ dy=\int_0^2\left(\frac{1}{2}x^2 + xy\right|_1^2\ dy=\int_0^2
\left(\frac{3}{2}+y\right)\ dy = \left(\frac{3}{2}y + \frac{1}{2}y^2\right|_0^2=5.\)

When the region of integration is not a rectangle, we can still use the same procedure. Let us evaluate the integral

\(\iint_R \left(x+y\right) \, dA,\)

where \(R\) is the region bounded by \(x=1\), \(y=0\), and \(y=x^2\).

Drawing horizontal and vertical lines through the region can help us find the limits. If we integrate \(dy\) then \(dx\), the interated
integral is

\(\int_0^1 \int_0^{x^2}(x+y)\ dy \ dx.\)

When we integrate \(dx\) then \(dy\), the integral is

\(\int_0^1 \int_\sqrt{y}^1 (x+y)\ dx \ dy\)

Most people prefer to do the first integral because both integrals have 0 as the lower limit. Do the integral of your choice, and check your solution below.

Show the solution

\(\int_0^1 \int_0^{x^2} \left(x+y\right) \, dy\ dx=
\int_0^1 \left(xy+\frac{1}{2}y^2\right|_0^{x^2} dx=
\int_0^1\left(x^3+\frac{1}{2}x^4\right)\, dx = \left(\frac{1}{4}x^4 + \frac{1}{10}x^5\right|_0^1=\frac{7}{20}\)

\(\int_0^1 \int_\sqrt{y}^{1} \left(x+y\right) \, dx\ dy=
\int_0^1 \left(\frac{1}{2}x^2 + xy\right|_\sqrt{y}^{1} dy=
\int_0^1\left(\frac{1}{2}+\frac{1}{2}y-y^{3/2}\right)\, dy = \left(\frac{1}{2}y+\frac{1}{4}y^2-\frac{2}{5}y^{5/2}\right|_0^1=\frac{7}{20}\)

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