Lecture2

# Lecture2 - Lecture 1: MATH 138-W12-003 January 6, 2012 I)...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: Lecture 1: MATH 138-W12-003 January 6, 2012 I) Integration by Parts, Section 7.1 ”I never met an integral I couldn’t integrate by parts” Recall that the product rule states, d dx [ f ( x ) g ( x )] = f ( x ) g ( x ) + f ( x ) g ( x ) . If we rearrange the terms by isolating the first term on the right and then integrate the equation we get a new and very useful technique for integrating functions, f ( x ) g ( x ) = d dx [ f ( x ) g ( x )]- f ( x ) g ( x ) , Z f ( x ) g ( x ) dx = Z d dx [ f ( x ) g ( x )] dx- Z f ( x ) g ( x ) dx, Z f ( x ) g ( x ) dx = f ( x ) g ( x )- Z f ( x ) g ( x ) dx. Therefore we have proven that, Z f ( x ) g ( x ) dx = f ( x ) g ( x )- Z f ( x ) g ( x ) dx . This last equation above is known and the formula for integration by parts . If we set u = f ( x ) and v = g ( x ) then it can be more compactly written as (using substitution), Z udv = uv- Z vdu . If we have bounds on the integral we do the same thing but now we have to evaluate the first term on the right at the two bounds,...
View Full Document

## This note was uploaded on 02/14/2012 for the course MATH 138 taught by Professor Anoymous during the Fall '07 term at Waterloo.

### Page1 / 2

Lecture2 - Lecture 1: MATH 138-W12-003 January 6, 2012 I)...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online