Iterated Integrals

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Regions of Type \(I\) and Type \(II\)

The most powerful tool for the evaluation of the double integrals is the Fubini's theorem. It works not for a general region \(R\) but for some special regions which we call Regions of type \(I\) or type \(II\).

Definition 1.

A plane region \(R\) is said to be of type \(I\) if it lies between the graphs of two continuous functions of \(x\) (Figure \(1\)), that is

\[R = \left\{ {\left( {x,y} \right)|\;a \le x \le b,\;\; p\left( x \right) \le y \le q\left( x \right)} \right\}.\]

Definition 2.

A plane region \(R\) is said to be of type \(II\) if it lies between the graphs of two continuous functions of \(y\) (Figure \(2\)), that is

\[R = \left\{ {\left( {x,y} \right)|\;u\left( y \right) \le x \le v\left( y \right),\;\; c \le y \le d} \right\}.\]

Fubini's Theorem

Let \(f\left( {x,y} \right)\) is a continuous function over a type \(I\) region \(R\) such that

\[R = \left\{ {\left( {x,y} \right)|\;a \le x \le b,\;\; p\left( x \right) \le y \le q\left( x \right)} \right\}.\]

Then the double integral of \(f\left( {x,y} \right)\) in this region is expressed in terms of the iterated integral:

\[\iint\limits_R {f\left( {x,y} \right)dA} = \int\limits_a^b {\int\limits_{p\left( x \right)}^{q\left( x \right)} {f\left( {x,y} \right)dydx} } .\]

For a region of type \(II\) we have the similar theorem:

If \(f\left( {x,y} \right)\) is a continuous function on a type \(II\) region \(R\) such that

\[R = \left\{ {\left( {x,y} \right)|\;u\left( y \right) \le x \le v\left( y \right),\;\; c \le y \le d} \right\}.\]

then

\[\iint\limits_R {f\left( {x,y} \right)dA} = \int\limits_c^d {\int\limits_{u\left( y \right)}^{v\left( y \right)} {f\left( {x,y} \right)dxdy} } .\]

Thus, the Fubini's theorem allows to calculate double integrals through the iterated ones. To evaluate an iterated integral, we first find the inner integral and then the outer integral.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Evaluate the iterated integral \[\int\limits_0^1 {\int\limits_1^2 {xydydx} }.\]

Example 2

Find the iterated integral \[\int\limits_0^1 {\int\limits_y^{{y^2}} {\left( {x + 2y} \right)dxdy} }.\]

Example 1.

Evaluate the iterated integral \[\int\limits_0^1 {\int\limits_1^2 {xydydx} }.\]

Solution.

We first evaluate the inner integral and then the outer integral:

\[\int\limits_0^1 {\int\limits_1^2 {xydydx} } = \int\limits_0^1 {\left[ {\int\limits_1^2 {xydy} } \right]dx} = \int\limits_0^1 {\left[ {x\left. {\left( {\frac{{{y^2}}}{2}} \right)} \right|_1^2} \right]dx} = \int\limits_0^1 {\frac{3}{2}dx} = \frac{3}{2}\left. {\left( {\frac{{{x^2}}}{2}} \right)} \right|_0^1 = \frac{3}{4}.\]

Example 2.

Find the iterated integral \[\int\limits_0^1 {\int\limits_y^{{y^2}} {\left( {x + 2y} \right)dxdy} }.\]

Solution.

Here we have the region of type \(II.\) Applying the Fubini's theorem we obtain

\[\int\limits_0^1 {\int\limits_y^{{y^2}} {\left( {x + 2y} \right)dxdy} } = \int\limits_0^1 {\left[ {\int\limits_y^{{y^2}} {\left( {x + 2y} \right)dx} } \right]dy} = \int\limits_0^1 {\left[ {\left. {\left( {\frac{{{x^2}}}{2} + 2yx} \right)} \right|_y^{{y^2}}} \right]dy} = \int\limits_0^1 {\left[ {\left( {\frac{{{y^4}}}{2} + 2{y^3}} \right) - \left( {\frac{{{y^2}}}{2} + 2{y^2}} \right)} \right]dy} = \int\limits_0^1 {\left[ {\frac{{{y^4}}}{2} + 2{y^3} - \frac{{5{y^2}}}{2}} \right]dy} = \left. {\left[ {\frac{{{y^5}}}{{10}} + \frac{{{y^4}}}{2} - \frac{{5{y^3}}}{6}} \right]} \right|_0^1 = \frac{1}{{10}} + \frac{1}{2} - \frac{5}{6} = - \frac{7}{{30}}.\]