Book Edition | 11th Edition |
Author(s) | Larson |
ISBN | 9781337275378 |
Publisher | Cengage Learning |
Subject | Math |
Explain what is meant by an iterated integral. How is it evaluated?
*To give you a brief description, iterated integral is just an integration of more than 1 variable, f(x, y). This is commonly used when getting an area using integration. It is evaluated by integrating first the inner portion of the equation. So here is the process on how to conduct iterated integration:
First, recall the formula of iterated integration
$∫∫f(x,y)dA=∫_{y_{1}}∫_{x_{1}}f(x,y)dxdy$
For this formula, we will first integrate the "inner" part of the equation.
$∫_{y_{1}}[∫_{x_{1}}f(x,y)dx]dy$
As you can see above, the function will be integrated with respect to x. This means that x is the variable and we will treat y as constant.
Second, evaluate the inner part of the equation.
$∫_{y_{1}}[f(x,y)]_{x_{1}}dy$
Third, after evaluating the inner part with respect to x, it is now time to integrate the function with respect to y. In this, x now serves as the constant while y is the variable.
*Let us have an example for us to understand how iterated integration is being done.
$∫_{0}∫_{1}x_{2}ydxdy$
In this problem, we need to integrate the inner part of the function first. Let us separate it for us to visualize more.
$∫_{0}∫_{1}x_{2}ydxdy=∫_{0}[∫_{1}x_{2}ydx]dy$
Integrate now with respect to x, treating y as constant.
$∫_{0}[∫_{1}x_{2}ydx]dy=∫_{0}[3x_{3}y ]_{1}dy$ = $∫_{0}[33_{3}y −31_{3}y ]dy=∫_{0}[326y ]dy$ = $326 ∫_{0}[y]dy$
Next, integrate with respect to y.
$326 ∫_{0}ydy=326 [2y_{2} ]_{0}=326 (22_{2} −20_{2} )=352 $
Iterated integral is just an integration of more than 1 variable, f(x, y). This is commonly used when getting an area using integration. It is evaluated by integrating first the inner portion of the equation.