Calculus II Notes 8.1

# Calculus II Notes 8.1 - Calculus II-Stewart Dr. Berg Spring...

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Calculus II- Stewart Dr. Berg Spring 2010 Page 1 8.1 8.1 Integration by Parts Recall that, by the product rule, d dx f ( x ) g ( x ) [ ] = f ( x ) g ( x ) + f ( x ) g ( x ) . Therefore f ( x ) g ( x ) = f ( x ) g ( x ) dx + f ( x ) g ( x ) dx . This yields the result known as integration by parts . Theorem f ( x ) g ( x ) dx = f ( x ) g ( x ) f ( x ) g ( x ) dx . Note: Using the variable substitutions u = f ( x ) and v = g ( x ) this becomes u dv = uv v du . The idea behind this technique is that, hopefully, v du is easier to integrate than u dv . Generally, the derivative of u should be simpler than u , and the antiderivative of dv should be no more complex than v . Example A (example 1 page 489) Evaluate x sin x dx . Solution : Let u = x and dv = sin x dx . Then du = dx and v = cos x . Therefore x sin x dx = x cos x ( ) cos x ( ) dx = x cos x + sin x + C . Example B (example 2 page 489) Evaluate ln x dx . Solution : Let u = ln x and dv = dx . Then du = 1 x dx and v = dx = x . Therefore ln x dx = x ln x x 1 x dx = x ln x dx = x ln x x + C Example C Evaluate x 2 cos x dx .

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## This note was uploaded on 02/28/2010 for the course M 56495 taught by Professor Berg during the Spring '10 term at University of Texas.

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Calculus II Notes 8.1 - Calculus II-Stewart Dr. Berg Spring...

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