Iterated Integrals
We are about to look at a method of evaluating double integrals over rectangles and more general domains without using the definition of a double integral, but before we do so, we will first need to learn what an iterated integral.
Suppose that
is a two variable real-valued function, and suppose that
is integrable for
and
, that is
is integrable over the rectangle:
![{\displaystyle {\begin{aligned}\quad R=\{(x,y)\in \mathbb {R} ^{2}:a\leq x\leq b,c\leq y\leq d\}=[a,b]\times [c,d]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8877a66f3d306d8d2886adbc19eb4e245b69647)
Just like partial differentiation with respect to a specific variable, we can also partial integrate with respect to a specific variable. For example,
means that we integrate
with respect to
while holding the variable
as fixed. When we evaluate this integral, we will obtain a function in terms of
only, and hence, we could then integrate the result from
to
with respect to
as:

The result above is what we call an iterated integral which we will define below.
- Definition: If
is a two variable real valued function, and if
is integrable over the rectangle
, then the Iterated Integral of
over
is

.
We will see later that iterated integrals need not be over rectangles but instead can be done over more general domains.
One important property about iterated integrals is that we can partially integrate
with respect to either variable
or
first, and then continue onward with integrating with respect to the second variable, that is:

However, it is important to note that sometimes partial integrating with respect to a certain variable first will be a much easier process.
Let's now look at some examples of evaluating iterated integrals.
Example 1
Evaluate the following iterated integral
. Over what rectangle is
being integrated?
When evaluated iterated integrals over rectangles, we always want to work from the inside out. Let's first evaluate the inside integral with respect to
while holding
as fixed.
![{\displaystyle {\begin{aligned}\quad \int _{1}^{3}x^{3}+xy^{2}\,dy=\left[x^{3}y+{\frac {xy^{3}}{3}}\right]_{1}^{3}=\left(3x^{3}+9x\right)-\left(x^{3}+{\frac {x}{3}}\right)=2x^{3}+{\frac {26x}{3}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/605633f21ef80de7b975c5c9d307fdf9046dd4c1)
Therefore we have that:

Evaluating this definite integral and we get that:
![{\displaystyle {\begin{aligned}\quad \int _{2}^{4}2x^{3}+{\frac {26x}{3}}\,dx=\left[{\frac {x^{4}}{2}}+{\frac {26x^{2}}{6}}\right]_{2}^{4}=\left(128+{\frac {416}{6}}\right)-\left(8+{\frac {104}{26}}\right)=120+52=172\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed776cb6b384e5816d3fa5bee45ff69a3063d3e4)
Therefore we have that:

In this particular example, we are integrating over the rectangle
.
Example 2
Evaluate the following iterated integral
. Over what rectangle is
being integrated?
We will first start by evaluating the inner integral while holding
as fixed:
![{\displaystyle {\begin{aligned}\quad \int _{\frac {\pi }{2}}^{\pi }\sin x\cos y\,dx=\left[-\cos x\cos y\right]_{0}^{\pi }=\left(\cos y\right)-\left(-\cos y\right)=2\cos y\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef6eaad13173332bc9558e70b8ec96e71a1dd9ee)
Therefore we have that:

Evaluating this definite integral and we get that:
![{\displaystyle {\begin{aligned}\quad \int _{0}^{\pi }2\cos y\,dy=\left[2\sin y\right]_{0}^{\pi }=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68ca81390e3d9a95b3acf25e09d97ec66c9ccf0f)
In this particular example, we are integrating over the square
.
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