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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:
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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 ƒ sec( e 4 x ) tan( e 4 x ) + 3 cos 2 x ƒ (-sin x ) (d) h ( θ ) = θ 2 tan( √ θ ) Solution: Use Product Rule ( u ƒ v ) ± = u ƒ v ± + v ƒ u ± , with u = θ 2 , v = tan( √ θ ) , u ± = 2 θ, and v ± = sec 2 ( √ θ ) ƒ 1 2 √ θ h ± ( θ ) = θ 2 ƒ sec 2 ( √ θ ) ƒ 1 2 √ θ + 2 θ ƒ tan( √ θ ) (e) f ( u ) = e-u 1 + u 2 Solution: Use Quotient Rule ‡ w v · ± = v ƒ w ±-w ƒ v ± v 2 , with w = e-u , v = 1 + u 2 , w ± =-e-u , and v ± = 2 u f ± ( u ) = (1 + u 2 ) ƒ (-e-u )-e-u ƒ (2 u ) (1 + u 2 ) 2 =-e-u (1 + u 2 + 2 u ) (1 + u 2 ) 2...
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