Evaluate the iterated integral.
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
Say what?!
Iterated Integrals Explained
Iteration means to perform an action or process repeatedly.
So, if we are asked to compute a double integral, then we will have to apply our integration strategies repeatedly.
These types of integrals are called iterated integrals, as they are the result of applying one-variable integration more than once.
Why?
Recall that the partial derivates of a function \(f\left( {x,y} \right)\) are computed by holding one of the variables fixed and differentiating with respect to the other variable.
Well, the same thing is happening with the antidifferentiation of a multivariable function. We integrate one variable at a time while holding the other variable constant and working inside-out.
\begin{equation}
\int_{c}^{d} \int_{a}^{b} f(x, y) d x d y=\int_{c}^{d}\left(\int_{a}^{b} f(x, y) d x\right) d y \text { or } \int_{a}^{b} \int_{c}^{d} f(x, y) d y d x=\int_{a}^{b}\left(\int_{c}^{d} f(x, y) d y\right) d x
\end{equation}
But now for the exciting part…
Fubini’s Theorem
We can switch the order of integration when calculating double iterated integrals over rectangular regions.
How is this possible?
Because we have fixed values for \(x\) and \(y\), the order of integration, \(dx\) or \(dy\), doesn’t matter, the same way it doesn’t matter if switch the order of multiplication.
This ability to switch the order of integration is known as Fubini’s Theorem, which allows the order of integration to be changed as long as the double integral yields a finite answer when the integrand is replaced by its absolute value.
Let’s prove it
Evaluate \(f\left( {x,y} \right) = 3x – {y^2}\) over the rectangular region \(R = \left[ {0,4} \right] \times \left[ { – 1,2} \right]\).
Since we have fixed values for \(x\) and \(y\) (i.e., \(0 \le x \le 4\) and \( – 1 \le y \le 2\)), it doesn’t matter in which order we integrate. So, let’s start with \(dxdy\).
\begin{equation}
\int_{-1}^{2} \int_{0}^{4}\left(3 x-y^{2}\right) d x d y
\end{equation}To compute the double integral, we start with the inside first and only integrate with respect to \(x\), keeping \(y\) constant.
\begin{equation}
\begin{aligned}
&\int_{-1}^{2}\left[\int_{0}^{4}\left(3 x-y^{2}\right) d x\right] d y \\
&\int_{0}^{4}\left(3 x-y^{2}\right) d x=\frac{3 x^{2}}{2}-\left.x y^{2}\right|_{0} ^{4}=\left(\frac{3}{2}(4)^{2}-(4) y^{2}\right)-\left(\frac{3}{2}(0)^{2}-(0) y^{2}\right)=24-4 y^{2} \\
&\int_{-1}^{2} \underbrace{\left[\int_{0}^{4}\left(3 x-y^{2}\right) d x\right]}_{24-4 y^{2}}
\end{aligned}
\end{equation}Now that we’ve completed the inner integral, it’s time to integrate the outer integral with respect to \(y\).
\begin{equation}
\int_{-1}^{2}\left(24-4 y^{2}\right) d y=24 y-\left.\frac{4 y^{3}}{3}\right|_{-1} ^{2}=\left(24(2)-\frac{4(2)^{3}}{3}\right)-\left(24(-1)-\frac{4(-1)^{3}}{3}\right)=60
\end{equation}Alright, so let’s now demonstrate that we get the same result when integrating in reverse order, \(dydx\)
\begin{equation}
\int_{0}^{4} \int_{-1}^{2}\left(3 x-y^{2}\right) d y d x
\end{equation}Like before, we will work inside-out and integrate the inner integral first by integrating with respect to y and keeping \(x\) constant.
\begin{equation}
\int_{0}^{4}\left[\int_{-1}^{2}\left(3 x-y^{2}\right) d y\right] d x
\end{equation}\begin{equation}
\int_{-1}^{2}\left(3 x-y^{2}\right) d y=3 x y-\left.\frac{y^{3}}{3}\right|_{-1} ^{2}=\left(3 x(2)-\frac{(2)^{3}}{3}\right)-\left(3 x(-1)-\frac{(-1)^{3}}{3}\right)=9 x-3
\end{equation}\begin{equation}
\int_{0}^{4} \underbrace{\left[\int_{-1}^{2}\left(3 x-y^{2}\right) d y\right]}_{9 x-3} d x
\end{equation}Now we will evaluate the outer definite integral with respect to \(x\).
\begin{equation}
\int_{0}^{4}(9 x-3) d x=\frac{9 x^{2}}{2}-\left.3 x\right|_{0} ^{4}=\left(\frac{9(4)^{2}}{2}-3(4)\right)-\left(\frac{9(0)^{2}}{2}-3(0)\right)=60
\end{equation}
See!
We get the same answer no matter if we integrate \(dxdy\) or \(dydx\)!
And we’ve just verified Fubini’s Theorem by showing how switching the order of integration yields the same result when integrating over a rectangular region!
Cool!
Together we will work through several examples in detail and learn some tricks to help us decide which order of integration is easiest to compute. We’ll also review our essential integration techniques like u-substitution, trig integration, and even integration by parts.
Let’s jump right in!
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