Feb 1st.pdf - Lecture Notes Winter 2019 MATA37 CALCULUS II FOR THE MATHEMATICAL SCIENCES LEC03 Feb 1st 2:00pm 3:00pm Instructor Email Office Office

# Feb 1st.pdf - Lecture Notes Winter 2019 MATA37 CALCULUS II...

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Lecture Notes Winter 2019 MATA37 - CALCULUS II FOR THE MATHEMATICAL SCIENCES LEC03 , Feb 1st, 2:00pm - 3:00pm Instructor: Dr. Kathleen Smith Email: [email protected] Office: IC458 Office Hours: TBA
LEC03 , Feb 1st Richard Hong 1 INDEFINITE PROOF 1 Indefinite Proof Prove: IF f and g 0 are cont on [a, b] THEN Z b a f ( g ( x )) g 0 ( x ) dx = Z g ( b ) g ( a ) f ( u ) du Where u = g ( x ) , du = g ( x ) dx Suppose that f and g 0 are cont on [a, b] Let F be an antiderivative of f on [a, b] Claim: F ( g ( x )) is an antiderivative of f(g(x)) f’(x) on [a, b] Check: Let x [ a, b ] be arbitrary, ( F ( g ( x ))) 0 = F 0 ( g ( x )) · g 0 ( x ) = f ( g ( x )) · g 0 ( x ) And so Z b a f ( g ( x )) g 0 ( x ) = F ( g ( x )) | b a By FTOC I Thus we have F ( g ( b )) - F ( g ( x )) Recall Z b a f ( g ( x )) g 0 ( x ) dx = Z g ( b ) g ( a ) f ( u ) du Z g ( b ) g ( a ) f ( u ) du = F ( u ) | g ( b ) g ( a ) = F ( g ( b )) - F ( g ( a )) By FTOC I QED Examples 1. Z 1 0 2 - x = Z u (1) u (0) u · ( - du ) = Z 1 2 u 1 2 · ( - du ) = Z 2 1 u 1 2 du = [ 2 3 u 1 2 ] 2 1 = 2 3 [1 - 8] let u = 2 - x 2. Z sin 5 ( x ) · cos

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Unformatted text preview: 3 ( x ) dx = Z sin 3 ( x ) · cos 2 ( x ) · cos( x ) dx = Z sin 5 ( x ) · (1-sin 2 ( x )) · cos( x ) dx let u = sin( x ) , du = cos( x ) dx Z sin 5 ( x ) · cos 3 ( x ) dx = Z u 5 (1-u 2 ) du = Z u 5-u 7 du = u 6 6-u 8 8 + C = sin 6 ( x ) 6-sin 8 ( x ) 8 + C For an exercise, do this with u = cos( x ) 3. Z p 3 + x 2 x 5 dx = Z p 3 + x 2 x 2 2 · x · dx Let u = 3 + x 2 , du 2 = x · dx, x 2 = u-3 Z p 3 + x 2 x 5 dx = Z √ u ( u-3) 2 du 2 = 1 2 Z √ u · ( u 2-6 u + 9) du = 1 2 Z u 5 2-6 u 3 2 + 9 u 1 2 du Page 1...
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