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Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon Math 16b Another integration by parts example Integrate Z ln ( x ) dx Set u = ln ( x ) and v = x so that du = dx x and dv = dx . Z ln ( x ) dx = Z u dv = u v Z v du = x ln ( x ) Z x 1 x dx = x ln ( x ) Z dx = x ln ( x ) x + C Thomas Scanlon Math 16b The Fundamental Theorem of Calculus, recalled If F ( x ) = f ( x ) , then Z b a f ( x ) dx = F ( b ) F ( a ) Thomas Scanlon Math 16b An example Evaluate Z 10 1 ( ln ( x )) 4 x dx Substitute u = ln ( x ) so that du = dx x . Z ( ln ( x )) 4 x dx = Z u 4 du = 1 5 u 5 + C = 1 5 ( ln ( x )) 5 + C Z 10 1 ( ln ( x )) 4 x dx = 1 5 ( ln ( 10 )) 5 1 5 ( ln ( 1 )) 5 = 1 5 ( ln ( 10 )) 5 Thomas Scanlon Math 16b Substituting the limits as well If we substitute u = g ( x ) so that R f ( x ) dx = R h ( u ) du , then Z b a f ( x ) dx = Z g ( b ) g ( a ) h ( u ) du Indeed, if H ( u ) = h ( u ) and f ( x ) = h ( g ( x )) g ( x ) , then ( H ◦ g ) ( x ) = H ( g ( x )) g ( x ) = h ( g ( x )) g ( x ) = f...
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 Fall '06
 Sarason
 Calculus, Fundamental Theorem Of Calculus, Integration By Parts, dx, Thomas Scanlon, Thomas Scanlon Math

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