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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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Lesson Transcript

Instructor:
*Gerald Lemay*

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we show how to evaluate a double integral using iterative integration. A special case is also presented which simplifies the calculations.

The word ''**iteration**'' means to do something repeatedly. This word is often used in math. In this lesson, we evaluate a double integral by using iteration. First, we'll do a single integral on one of the variables and then repeat with a single integral on the other variable. Rather than repeat this information, let's do an example.

Imagine needing to find the solution to the double integral

We have two variables, *x* and *y*. Let's integrate on *x* first. To be clear, we can use parentheses:

The integral inside the parentheses is called the **inner integral**. We need to evaluate the integral of *y**x*2 from *x* = 2 to *x* = 3.

This is an integration on *x*. What about the *y* appearing in *y**x*2? In this integration on *x*, we treat *y* as if it were a constant. So, *y**x*2 is like *c**x*2 where *c* is a constant.

The integration of *c**x*2 is *c**x*3/3. So, when we integrate on *x*, the integral of *y**x*2 is *y**x*3/3.

Evaluating with the upper and lower limits:

Simplifying, we get 19*y*/3 as the answer for the inner integral. Next, we do the outer integral. We evaluate:

Do you see how the integral within the parentheses has been replaced with 19*y*/3?

Now, we iterate (i.e, repeat) by integrating over *y*:

Simplifying, we get 47.5 as the final answer.

The double integral was computed using an ''iterated integral.'' We started with an integration on *x*. What if we had started with an integration on *y*?

Then, the double integral would be written as:

Do you see how the work is clarified by using parentheses and by labeling the lower limit of the integrals?

The integration on *y* becomes:

Now, the variable is *y* while *x* is the constant. Continuing:

Now, the outer integral is over *x*:

Do you see how 15 divided by 3 produced 5?

Simplifying further, we get 47.5. Same as before. We can choose to iterate *x* then *y* or *y* then *x*. The result will be the same.

Something special happens when the function of *x* and *y*, f(*x*,*y*), can be written as the product of a function on *x*, g(*x*), multiplied with a function on *y*, h(*y*). Then, the iterative integral becomes the product of two integrals. This is really a special situation, but our example is just such a case. That's because we have f(*x*,*y*) = *y**x*2 which is the same as g(*x*)h(*y*) where g(*x*) = x2 and h(*y*) = *y*. What we are saying is:

Here are the details:

Let's check:

and

Thus:

Which is the same result as before.

Does this always happen? Yes, if the function being integrated can be written as a function of *x* multiplied by a function of *y*.

A double integral can be evaluated by repeated use of a single integral. This repetition is called an **iterative integral**. The variable integrated first can be written as an integral inside parentheses, called the **inner integral**. The remaining variable is integrated as the outer integral. When the function being integrated can be written as a function of *x* multiplied by a function of *y*, the iterated integral simplifies to the product of the integral of these separate functions. That is, if f(*x*,*y*) = g(*x*)h(*y*), then:

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GRE Math: Study Guide & Test Prep27 chapters | 182 lessons | 16 flashcard sets

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