# Iterated Integrals

## Trigonometry # Iterated Integrals

## 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

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