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# Class Note for MATH 250A with Professor Lega at UA

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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 13 views.

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Date Created: 02/06/15
Calculus and Differential Equations I MATH 250 A Methods of integration I Methods of integration Calculus and D rentiaI Equations Two methods of integration 9 Integration by substitution is a method based on the chain rule fgx gX dx fgx dx fgX C 0 Examples Find 2x1 ex2 ex dX and 4 coilLb dx 1 X 0 Integration by parts is a method based on the product rule ux VX dx ux vx iux vX dx 9 Examples Find x sinhx dX and Int dt Methods of integration Calculus and Differential Equations 7 method 0 use 7 0 When to make a substitution roughly speaking if the integrand involves a function of a complicated expression it is a good idea to define this complicated expression as a new variable a When to use integration by parts this method will work if the integral of LIX vX is somewhat simpler than that of uX v x Otherwise there is no reason to use integration by parts Note that sometimes one has to use integration by parts more than once 0 How would you evaluate the integral X3 sinX dx 0 Integration by parts 9 Integration by substitution 9 None of the above in I continued Which method 0 use X2 H X2 dx 0 How would you evaluate the integral 0 Integration by parts 9 Integration by substitution 9 None of the above X2 H Xs dx 0 How would you evaluate the integral 0 Integration by parts 9 Integration by substitution 9 None of the above 0 How would you evaluate the integral X4 expx3 dx 0 Integration by parts 9 Integration by substitution 9 None of the above M aims of integr Calculus and Differential Equations Methods of integration Calculus and Dif ntiaI Equations Some of your questions a If one has to make a substitution to evaluate a definite integral what should one do about the limits of integration 0 Leave them unchanged 9 Recalculate them in terms of the new variable 9 First find an antiderivative and then use the fundamental theorem of Calculus to evaluate the definite integral 0 Either 1 or 3 9 Either 2 or 3 3 None of the above 0 Can you find an example of an integral that can be evaluated both by substitution and by integration by parts 0 Yes 9 No Methods of I in Calculus and Dilf ntial Equationsl Some of your questions continued 0 If we cannot evaluate an integral does it mean that the integrand does not have an antiderivative 0 Yes 9 No a If one tries to use integration by parts in a situation where the integral can only be evaluated by substitution is the method of integration by parts going to give a wrong answer 0 Yes 9 No 0 If one has to use integration by parts repeatedly to evaluate a definite integral is it better to evaluate the brackets as one goes along or to find an antiderivative first and then evaluate the definite integral 39 of integration Calculus and Differential Equations

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