q1-sol-1042-sp08

q1-sol-1042-sp08 - u = e 4 x , u = 4 e 4 x and Power Chain...

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Calculus II—1042 Solution Quiz 1 1. Compute the follwing integrals (a) Z ± 1 3 + 2 5 x - x 2 dx Solution: Z ± 1 3 + 2 5 x - x 2 dx -→ Z ± 1 3 + 2 ƒ x - 1 / 5 - x 2 dx = 1 3 ƒ x + 2 ƒ x 4 / 5 4 / 5 - 1 2 ƒ x 2 2 + C = x 3 + 5 x 4 / 5 2 - x 2 4 + C (b) Z e t/ 3 - e - t · dt Solution: Z e t/ 3 - e - t · dt = 1 1 / 3 e t/ 3 - 1 - 1 e - t + C = 3e t/ 3 + e - t + C (c) Z ( 5 sec 2 y + csc y cot y ) dy Solution: Z ( 5 sec 2 y + csc y cot y ) dy = 5tan y - csc y + C (d) Z ± 2 1 + u 2 + 3 sin(5 u ) du Solution: Z ± 2 1 + u 2 + 3 sin(5 u ) du = 2 ƒ tan - 1 u + 3 ƒ ± - 1 5 sin(5 u ) + C = 2tan - 1 u - 3 sin(5 u ) 5 + C 2. Solve the initial value problem: f ± ( x ) = 2 x 4 + 1 2 x , f ( - 1) = 3 5 Solution: Take anti-derivative of both sides. f ( x ) = Z f ± ( x ) dx = Z ± 2 x 4 + 1 2 x dx = 2 ƒ x 5 5 + 1 2 ƒ ln | x | + C . Now plug in x = - 1 , f ( x ) = 3 / 5. We have 3 / 5 = 2( - 1) 5 + 1 2 ln | - 1 | + C = 3 / 5 = - 2 / 5 + C = C = 1 f ( x ) = 2 x 5 5 + ln | x | 2 + 1 3. Compute the derivative with respect to an appropriate variable. (a) y = sin - 1 ( x 2 ) Solution: Use the chain rule with u = x 2 , u ± = 2 x y ± = 1 p 1 - ( x 2 ) 2 ƒ 2 x = 2 x 1 - x 4 (b) g ( t ) = ln ± e t ( t 3 - 1) 4 Solution: First use the Logrithm’s Rules to simplify then differentiate. g ( t ) = ln ± e t ( t 3 - 1) 4 = ln( e t ) - ln ( ( t 3 - 1) 4 ) = t - 4 ln( t 3 - 1), use the Chain Rule for the sec- ond term with u = t 3 - 1 , u ± = 3 t 2 g ± ( t ) = 1- 12 t 2 t 3 - 1
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(c) y = sec( e 4 x ) + cos 3 ( x ) Solution: Use Chain Rule for the first term with
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Unformatted text preview: u = e 4 x , u = 4 e 4 x and Power Chain Rule for the second term as cos 3 ( x ) = (cos x ) 3 y = 4e 4 x sec( e 4 x ) tan( e 4 x ) + 3 cos 2 x (-sin x ) (d) h ( ) = 2 tan( ) Solution: Use Product Rule ( u v ) = u v + v u , with u = 2 , v = tan( ) , u = 2 , and v = sec 2 ( ) 1 2 h ( ) = 2 sec 2 ( ) 1 2 + 2 tan( ) (e) f ( u ) = e-u 1 + u 2 Solution: Use Quotient Rule w v = v w -w v v 2 , with w = e-u , v = 1 + u 2 , w =-e-u , and v = 2 u f ( u ) = (1 + u 2 ) (-e-u )-e-u (2 u ) (1 + u 2 ) 2 =-e-u (1 + u 2 + 2 u ) (1 + u 2 ) 2...
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This homework help was uploaded on 04/18/2008 for the course MATH 1042 taught by Professor Dr.z during the Spring '08 term at Temple.

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q1-sol-1042-sp08 - u = e 4 x , u = 4 e 4 x and Power Chain...

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