VIEWS: 13 PAGES: 18 POSTED ON: 6/23/2012 Public Domain
Integration by Substitution Antidifferentiation of a Composite Function 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 f g x g '( x) dx F ( g ( x) C Exploration • Recognizing Patterns. Discover the rule using the exploration Recognizing the f’(g(x)) g’(x) Pattern • Evaluate x 1 2 x dx 3 2 g '( x) 2 x and f ( g ( x)) g x 3 g ( x) 4 C ( x 2 1 C 4 4 Recognizing the f’(g(x)) g’(x) Pattern • Evaluate 1 1 2 cos d 1 1 Letting g ( x ) and g '( x ) 2 1 1 1 f ( g ( x)) cos cos 2 d 1 sin C Multiplying and Dividing by a Constant • Evaluate x 4x 3 dx 2 3 Rewrite the integral as 4x 3 x dx 2 3 4 x 3 8 x dx 1 2 3 8 1 4 x 3 4 x 3 C 2 4 2 4 C 8 4 32 Evaluating an Integral t 2t2 t dt 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 x 2 x 1 dx • 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 u 1 x Next find the derivative of this function 2 du dx Then substitute the variables into the given integral 2 u 1 12 du 2 u 2 Now symplify Integration by Substitution 1 u 1 u 2 du 1 4 1 3 u u 2 du 4 2 1 5 3 1 u 2 u 2 1 52 1 32 3 C u u C 4 5 10 6 2 2 Now substitute (2 x 1) back in for u. Integration by Substitution 1 1 2 x 1 2 x 1 2 C 5 3 2 10 6 • 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 2 x 1 2 3 2 x 1 5 1 3 30 1 1 2 x 1 6 x 3 5 2 x 1 2 6 x 8 3 3 2 30 30 Integrating Trig Functions • Remember that you need the derivative of the trig function and the function. Evaluate sin 2 3x cos 3x dx 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 g ( x) n 1 g ( x) g '( x) dx C , n 1 n n 1 Equivalently, if u g ( x) then n1 u u du n 1 C n 1 n Definite Integrals 3 x( x 1) dx 2 3 Evaluate 0 x 1 2 x dx 1 3 2 3 2 0 3 x 2 1 x 2 1 3 1 02 1 3 4 4 2 4 1 2 4 8 8 8 0 0 10, 000 1 9999 8 8 8 Integration of Even and Odd Functions • Let f be integrable on the closed interval, [a, –a ], • 1. If f is an even function, then a a a f ( x) dx 2 f ( x) dx 0 • 2. If f is an odd function, then a a f ( x) dx 0.