## Multivariate Calculus for Actuarial Science, Part I

### This *very* brief introduction to multivariate calculus is designed to help you develop the technical skills needed to pass
Exam P. For a thorough treatment, including the theory behind the material and many more examples, consult your favorite
calculus textbook.

This tutorial will proceed in three steps. First, you will learn how to evaluate a double integral. After that, you will learn how to set
up double integrals. Finally, you will learn how to reverse the order of integration in a double integral. At that point you will have
acquired some of the basic skills necessary to study probability when more than one variable is involved.

What does an expression like this mean?

\(\int_0^1 \int_2^x \left(x+y^2\right) \, dy\ dx\)

These are sometimes called *iterated integrals* because we integrate once with respect to one variable, then again
with respect to a different variable. We could also write the expression this way.

\(\int_0^1 \left( \int_2^x \left(x+y^2\right) \, dy \right) dx\)

We will evaluate the inner integral first, using \(y\) as our variable, treating \(x\) as a constant, then we integrate the resulting expression with respect to \(x\).

\(\int_0^1 \left( \int_2^x \left(x+y^2\right) \, dy \right) dx=
\int_0^1 \left(\left(xy+\frac{1}{3}y^3\right|_2^x\right) dx=
\int_0^1 \left(\frac{x^3}{3}+x^2-2 x-\frac{8}{3}\right) dx=
\left(\frac{x^4}{12}+\frac{x^3}{3}-x^2-\frac{8 x}{3}\right|_0^1=
-13/4
\)

Notice that after we integrated \(dy\), there were no \(y\)'s remaining. The variable \(x\) could appear in the limits of integration,
but not \(y\). Let's look at another example.

\(\int_0^3 \int_0^{y}\left(2x+e^{y^2}\right) \, dx dy=
\int_0^3 \left(x^2+xe^{y^2}\right|_0^{y} dy=
\int_0^3 \left(y^2+y\, e^{y^2}\right) dy=
\left(\frac{y^3}{3}+\frac{e^{y^2}}{2}\right|_0^3=
\frac{1}{2} \left(17+e^9\right)
\)

Now it's your turn. Evaluate the integral

\(\int_0^1 \int_0^2 \left(x+y+3\right) \, dy dx\)

then click on the box below to see the solution.

Show the solution

\(\int_0^1 \int_0^2 \left(x+y+3\right) \, dy dx=
\int_0^1 \left(xy+\frac{y^2}{2}+3y\right|_0^2 dx=
\int_0^1(2x+8)\, dx = \left(x^2 + 8x\right|_0^1=9\)

Move on to the next page