Calculus 2 - Assignment 1 Solutions

# Calculus 2 - Assignment 1 Solutions - t = 2 and t = 7 years...

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MAT1332 Spring/Summer 2009 Assignment 1 Solutions 1. Z 3 - 2 ( x 4 - 6 x 3 + x ) dx = x 5 5 - 3 x 4 2 + x 2 2 ·ﬂ 3 - 2 = 243 5 - 243 2 + 9 2 + 32 5 + 24 - 2 = - 40 2. Z π 4 0 ( - 96 x 2 + 32tan( x )) dx = 32 x 3 - 32ln | cos( x ) | ·ﬂ π /4 0 = - π 3 2 - 32ln | cos( π 4 ) |+ 32ln | cos(0) | =- π 3 2 - 32ln | 1 p 2 | = - π 3 2 + 32 2 ln | 2 | = - π 3 2 + 16ln | 2 | 3. First we calculate the indeﬁnite integral using the substitution u : = π 2 (3 x - 1) to give dx = 3 du 2 π and Z sin ( π 2 (3 x - 1) ) dx = Z sin( u ) 3 2 π du = - 3 2 π cos( u ) + = - 3 2 π cos ( π 2 (3 x - 1) ) + C . Thus, Z 2 - 4 sin ( π 2 (3 x - 1) ) dx = - 3 2 π cos ( π 2 (3 x - 1) ) ·ﬂ 2 - 4 = - 2 3 π cos( 5 π 2 ) + 2 3 π cos( - 13 π 2 ) = 0 4. Integration by parts gives Z 4 x sin( x ) dx = 4 ( - x cos( x ) + Z cos( x ) dx ) = 4 ( - x cos( x ) + sin( x ) ) + Cc . Therefore, Z π 2 0 4 x sin( x ) dx = 4 ( - x cos( x ) + sin( x ) ) ·ﬂ π 2 0 = - 2 π cos( π 2 ) + 4sin( π 2 ) - 4sin(0) = 4. 5. Substitution u : = 3 x - 1 gives dx = 1 3 du and Z 1 3 x - 1 dx = 1 3 Z 1 u du = 1 3 ln | 3 x - 1 |+ C . 6. Total growth of the ﬁsh between
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Unformatted text preview: t = 2 and t = 7 years is given by the deﬁnite integral Z 7 2 7.43 e-0.07 t dt = ‡-106.143 e-0.07 t ) ﬂ ﬂ ﬂ 7 2 = 27.2503. I.e., the ﬁsh grew 27.2503 centimeters between years 2 and 7. 7. The total amount of chemical produced between t = 2 and t = 7 is given by the deﬁnite integral Z 7 2 6 e-3.0 t dt = -2 e-3.0 t ﬂ ﬂ ﬂ 7 2 = 0.0050. I.e., in the time between minute 2 and minute 7 there were 0.0050 moles produced. 1...
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## This note was uploaded on 01/16/2011 for the course MAT 1332 taught by Professor Munteanu during the Fall '07 term at University of Ottawa.

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