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Integration by Substitution Lesson 5.5 Substitution with Indefinite Integration • This is the “backwards” version of the chain rule • Recall … d 2 x 4x 7 dx 5 5 x 4x 7 2 2x 4 4 • Then … 5 x 2 4x 7 2 x 4 dx x 4 2 5 4x 7 C Substitution with Indefinite Integration f ( x)dx • In general we look at the f(x) and “split” it into a g(u) and a du/dx du f ( x) g (u ) dx • So that … du g ( u ) dx G ( u ) C dx Substitution with Indefinite Integration • Note the parts of the integral from our example 5 x 2 4x 7 2 x 4 dx x 4 2 5 4x 7 C du g ( u ) dx G ( u ) C dx Example • Try this … what is the g(u)? what is the du/dx? (4 x 5) dx • We have a problem … 2 Where is the 4 which we need? Example • We can use one of the properties of integrals c f ( x)dx c f ( x)dx Where did the 1/3 come from? Why is this inside now anda 3? a • We will insert a factor of 4 factor of ¼ outside to balance the result 1 1 1 2 3 (4 x 5) 4 (4 x 5 ) dx 4 3 4 Can You Tell? • Which one needs substitution for integration? x 3 x 5 dx 2 x 3 x 5 dx 2 5 • Go ahead and do the integration. Try Another … 3t 1 dt sin x cos x dx 3 Assignment A • Lesson 5.5 • Page 340 • Problems: 1 – 33 EOO 49 – 77 EOO Change of Variables • We completely rewrite the integral in terms of u and du x 2 x 3 dx • Example: • So u = 2x + 3 and du = 2x dx • But we have an x in the integrand So we solve for x in terms of u u 3 x 2 Change of Variables • We end up with 1 2 1 x 2 x 3 dx u 3 u du 2 1 2 • It remains to distribute the u and proceed with the integration • Do not forget to "un-substitute" What About Definite Integrals 2 • Consider a variation 3t 1 dt of integral from previous slide 1 • One option is to change the limits u = 3t - 1 Then when t = 1, u = 2 when t = 2, u = 5 Resulting integral 5 1 2 1 u du 32 What About Definite Integrals • Also possible to "un-substitute" and use the original limits 1 2 3 2 1 1 2 2 u du u 3 t 1 3 3 3 9 3 2 2 1 Integration of Even & Odd Functions • Recall that for an even function f ( x) f ( x) The function is symmetric about the y-axis a • Thus a a f ( x) dx 2 f ( x ) dx 0 • An odd function has f ( x) f ( x) The function is symmetric about the orgin a • Thus a f ( x) dx 0 Assignment B • Lesson 5.5 • Page 341 • Problems: 87 - 109 EOO 117 – 132 EOO