Substitution Michael Freeze MAT 151 UNC Wilmington Summer 2013 1 - PowerPoint PPT Presentation

Substitution Michael Freeze MAT 151 UNC Wilmington Summer 2013 1 / 17

Section 7.2 :: Substitution 2 / 17

Substitution Form of the Integral � [ f ( x )] n f ′ ( x ) dx , n � = − 1 Result of Substituting u = f ( x ) n + 1 + C = [ f ( x )] n +1 u n du = u n +1 � + C n + 1 3 / 17

Substitution Form of the Integral � f ′ ( x ) f ( x ) dx Result of Substituting u = f ( x ) � 1 u du = ln | u | + C = ln | f ( x ) | + C 4 / 17

Substitution Form of the Integral � e f ( x ) f ′ ( x ) dx Result of Substituting u = f ( x ) � e u du = e u + C = e f ( x ) + C 5 / 17

Substitution Method Often we can choose u to be one of the following: 1 the quantity under a root or raised to a power 2 the quantity in the denominator 3 the exponent on e 6 / 17

Basic Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � (2 x + 3) 3 · 2 dx 7 / 17

Basic Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � 5 · (5 x − 9) 2 dx 8 / 17

Basic Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � ( − 4 x + 1) 7 dx 9 / 17

Basic Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � � � 4 · 2 x dx 3 x 2 − 5 10 / 17

Intermediate Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � x + 1 ( x 2 + 2 x − 4) 4 dx 11 / 17

Intermediate Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � x 2 � 2 x 3 + 7 dx 12 / 17

Intermediate Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. x 2 � √ dx 2 x 3 + 7 3 13 / 17

Intermediate Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � x e 3 x 2 − 5 dx 14 / 17

Advanced Integration by Substitution Examples General Power Rule for Integration For any real number n � = − 1, [ f ( x )] n f ′ ( x ) dx = [ f ( x )] n +1 � + C . n + 1 Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � (1 + ln x ) 2 dx x 15 / 17

Advanced Integration by Substitution Examples Recall that � f ′ ( x ) f ( x ) dx = ln | f ( x ) | + C Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. � x + 1 x 2 + 2 x − 4 16 / 17

Advanced Integration by Substitution Examples Recall that � f ′ ( x ) f ( x ) dx = ln | f ( x ) | + C Here, u = f ( x ) and du = f ′ ( x ) dx . Evaluate the indefinite integral. e 2 x � e 2 x + 3 dx 17 / 17