# 7.2 - MATH 1081  Wednesday, April 6   ...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 1081  Wednesday, April 6    Chapter 7 – Section 2    SUBSTITUTION  Homework #11 (due 4/11):              Clicker Check­in:  Choose any letter to check in now.      Section 7.1 #16, 20, 54, 60  Section 7.2 #18, 24, 32, 40, 42  When we evaluate the integral  ∫ 2 x x − 5 dx , we do have the  2 6 ( ) option to multiply the function out and then evaluate term by  term, but this would require some fairly lengthy algebra.      6 7 12 2 Instead, if we consider that  ∫ 2 x x − 5 dx = x − 5 + C ,  7 what happened to the  2 x  part of the original integrand?  ( ) ( ) The method of substitution can be used to simplify an  integrand.  In particular, when we have an integrand that is the  result if the chain rule for differentiation, substitution will  allow us to detect and remove the derivative of the “inside”  function to make a simpler integral to evaluate.      Recall the chain rule for differentiation:            D ⎡ f ( g ( x ) ) ⎤ = f ' ( g ( x ) ) ⋅ g ' ( x ) .  ⎣ ⎦ So,          ∫ f ' ( g ( x )) ⋅ g ' ( x ) dx = f ( g ( x )) + C .  To see how this works, let’s look again at  ∫ 2 x x − 5 dx .  2 6 ( )   6 2 If we let  u = x − 5 , then we have  ∫ 2 x ( u ) dx .  Already this looks  simpler, but there are now 2 variables, which is a problem.    du 2 However, notice that if  u = x − 5 , then  = 2 x  or  du = 2 x dx .  dx   Substituting this in as well, we now have  ∫ u 6 du  which is very  simple to evaluate.     7 17 12 So, we have  ∫ 2 x x − 5 dx = ∫ u du = u + C = x − 5 + C .  7 7 2 6 6 ( ) ( ) If we think of the chain rule as  “FIRST take the derivative of  the outside function, THEN take the derivative of what’s  inside”, then this technique of integration by substitution  “FIRST undoes the derivative of what’s inside, THEN undoes  the derivative of the outside function”.      The key is to look for an “inside” function and its derivative.    Example:  Evaluate the indefinite integral.    6x2       1.   ∫ 3/ 2 dx   3 2x + 7     ey       2.   ∫ dy   y             3.   ∫ ( 3x − 9 ) x 2 − 6 x dx   ( )   Example:  Evaluate the integral.                                  1.   ∫ p ( p + 1) dp   5 −4 x 2.   ∫ 2 dx   x +3 −4 x 3.  ∫ dx   x+3 Clicker Check­out:  Press any letter to check out now.    Tomorrow in recitation:  Workshop    Next time – Section 7.3    ...
View Full Document

## This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

Ask a homework question - tutors are online