## 7.5: Strategy for Integration

Over Integrals Served. Right click on any integral to view in mathml.

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• 7.5: Strategy for Integration!

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The next topic that we should discuss here is the integration variable used in the integral.

## Integrals Tutorial

For instance,. Changing the integration variable in the integral simply changes the variable in the answer. This is more important than we might realize at this point. Another use of the differential at the end of integral is to tell us what variable we are integrating with respect to. However, if you are on a degree track that will take you into multi-variable calculus this will be very important at that stage since there will be more than one variable in the problem.

You need to get into the habit of writing the correct differential at the end of the integral so when it becomes important in those classes you will already be in the habit of writing it down. The second integral is also fairly simple, but we need to be careful. The second integral is then,. So, it may seem silly to always put in the dx , but it is a vital bit of notation that can cause us to get the incorrect answer if we neglect to put it in. See the Proof of Various Integral Formulas section of the Extras chapter to see the proof of this property.

Notice that when we worked the first example above we used the first and third property in the discussion. We integrated each term individually, put any constants back in and then put everything back together with the appropriate sign.

go here Not listed in the properties above were integrals of products and quotients. The reason for this is simple.

Just like with derivatives each of the following will NOT work. With derivatives we had a product rule and a quotient rule to deal with these cases. However, with integrals there are no such rules. When faced with a product and quotient in an integral we will have a variety of ways of dealing with it depending on just what the integrand is.

There is one final topic to be discussed briefly in this section. We can now answer this question easily with an indefinite integral. This method places the burden of calculation squarely on finding an antiderivative F of f.