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Lecture

Riemann Integration II.pdf

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Department
Statistics
Course
STAT312
Professor
Douglas Wiens
Semester
Fall

Description
129 26. Riemann Integration II • Example: Recall Taylor’s Theorem with the re- mainder in integral form. Suppose we wish to approximate the cube root of = 9 in terms of that of = 8; we will do this by expanding ( ) = 1 3 around . How close do we get if we stop after the quadratic term? ( )2 ( ) = ( ) + 0( )( ) + 00( ) 2! 1 Z 2 000 + ( ) ( ) 2! Here ( ) = 2, 0( ) = 1 12, 0( ) = 1 144: 1 3 1 1 9 2 + = 2 07986 12 288 With a calculator, 39 = 2 08008 with an error of 00022; the estimRte of the error using Taylor’s Theorem is = 1 8 (9 )2 00( ) , with (lab 2 question) 5 | | = 00024 (26.1) 20 736 130 • Improper Riemann integrals, in which one or both endpoints are innite, or at which is unbounded, are dened by taking appropriate limits: Z Z ( ) = lim ( ) Z Z Z ( ) = ( ) + ( ) for any Z Z ( ) = lim ( ) if ( ) = ± 0 • An application of the Fundamental Theorem of Calculus is the formula for integration by parts. If are di erentiable, and 0 , 0are integrable, then Z [ ( ) ( )]0 = ( ) ( ) ( ) ( ) and also Z h 0 0 i = ( ) ( ) + ( ) ) ; hence Z Z 0 0 ( ) ( ) = ( ) ( ) ( ) ( ) ( ) ( ) 131 R • Example. Dene ( ) = 1 , ( 0 0), the Gamma integral. — Existence? Here is an outline: If 1 we R need only show that ( ) = 0 1 has a nite limit as ; since ( ) is in- creasing it is enough for it to be bounded. But for large enough we have that 1 2 , whose integral is bounded. If 1 ... — Evaluation: Z à ! 0 ( ) = 0 à ! ¯ Z à ! ¯ = ¯
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