Integration by Substitution Antidifferentiation of a Composite Function.

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Presentation transcript:

Integration by Substitution Antidifferentiation of a Composite Function

Let g be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then

Exploration Recognizing Patterns. Discover the rule using the exploration

Recognizing the f’(g(x)) g’(x) Pattern Evaluate

Recognizing the f’(g(x)) g’(x) Pattern Evaluate

Multiplying and Dividing by a Constant Evaluate

Evaluating an Integral We presently have no division rule for integrals. As a matter of fact there will never be a rule that involves division. Suggestions?

Change of Variables We are allowed to multiply any integral by a constant, but we can never multiply by a variable. What can we do if there is an extra variable in the given equation?

Integration by Substitution Evaluate There is an extra x in this integrand. In order to solve this equation, we will let u = 2x – 1.

Integration by Substitution Now solve this equation for x

Integration by Substitution

This is a correct answer, but it can be simplified a little further. If the problem is not a multiple choice question on the test or quiz, this answer is completed. If it is a multiple choice question, it will be given as

Integration by Substitution

Integrating Trig Functions Remember that you need the derivative of the trig function and the function.

Guidelines for Making a Change of Variable 1. Choose a substitution u = g(x). Usually, it is best to choose the inner part of a composite function, such as a quantity raised to a power. 2.Compute du = g’(x) dx. 3.Rewrite the integral in terms of the variable u. 4.Evaluate the resulting integral in terms of u. 5.Replace u by g(x) to obtain an antiderivative in terms of x. 6.Check your answer by differentiating.

General Power Rule for Integration If g is a differentiable function of x, then

Definite Integrals

Integration of Even and Odd Functions Let f be integrable on the closed interval, [a, –a ], 1. If f is an even function, then 2. If f is an odd function, then