Section5_3_review

Section5_3_review - Section 5.3: The Fundamental Theorem of...

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Unformatted text preview: Section 5.3: The Fundamental Theorem of Calculus Let f be continuous on a , b [ ] . If F is any antiderivative of f a , b [ ] , then f t ( 29 dt a b ∫ = F t ( 29 a b = F b ( 29- F a ( 29 . Note that a constant added to the antiderivative F will cancel, so it is not necessary to include a constant. Example. Evaluate x 4 dx 1 3 ∫ . Solution. x 4 dx 1 3 ∫ = x 5 5 1 3 = 3 5 5- 1 5 5 = 3 5- 1 5 = 243- 1 5 = 242 5 Example. Evaluate 6 v- 5 ( 29 dv 4 10 ∫ . Solution. 6 v- 5 ( 29 dv 4 10 ∫ = 6 v 2 2- 5 v 4 10 = 3 10 ( 29 2- 5 10 ( 29 - 3 4 ( 29 2- 5 4 ( 29 ( 29 = 300- 50- 48 + 20 = 222 Example. Evaluate 3sin xdx- π 2 π 2 ∫ . Solution. 3sin xdx- π 2 π 2 ∫ = 3- cos x ( 29 ]- π 2 π 2 = - 3cos π 2-- 3cos- π 2 = - 3 0 ( 29 + 3 0 ( 29 = Example. Evaluate r 2 + π r 3 ( 29 dr 1 2 ∫ . Solution....
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This note was uploaded on 01/12/2010 for the course MATH 44245 taught by Professor Famiglietti during the Fall '07 term at UC Irvine.

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Section5_3_review - Section 5.3: The Fundamental Theorem of...

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