Report

→ Definite Integral in terms of x from (lower bound) x=a to (upper bound) x=b Use the substitution u = f (x) d u = f ' (x) d x c = f (a) (lower bound) d = f (b) (upper bound) to get → =D Definite Integral in terms of u From u=c to u=d Then, INTEGRATE and use F.T.C to get =D Note: D is a real number Copyright © 2008 Pearson Education, Inc. Slide 7-1 EXAMPLE e² Evaluate I = ∫1 (-6) [x ( 7 - 3 ln x )] -1 d x. Let u = 7 - 3 ln x , so that d u = - 3 x -1 d x , [and thus x -1 d x = d u / (-3) x = 1, then u = 7 - 3 ln (1) = 7 – 3(0) = 7 x = e², then u = 7 - 3 ln (e²) = 7 – 3(2) = 1 Then with these substitutions, we get: 1 I=∫ (-6) u-1 d u / (-3) 7 1 lower bd in x < upper bd in x i.e. 1 < e² lower bd in u > upper bd in u i.e. 7 > 1 1 = ∫7 (-6)/(-3) u d u = 2 ∫7 u -1 d u 1 = 2 ln |u| | 7 = 2 [ ln (1) – ln (7) ] = - 2 ln (7) ≈ - 3. 892 -1 Note: The definite integral in terms of u is equal to a real number. And thus the definite integral in terms of x is also equal to that real number. The value of a definite integral that is not representing anEducation, area, Inc. does not need to be positive. Slide 7-2 Copyright © 2008 Pearson