Solutions2

# Solutions2 - = 3 , B =-3 , C = 2 ⇒ Z 5 x + 1 ( x-1)( x 2...

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MATH 2007B Test 2 Solutions October 24, 2011 [Marks] 1. Evaluate Z dx (9 - x 2 ) 3 / 2 . [6] Solution: Let x = 3 sin( t ) , - π 2 t π 2 . Then dx dt = 3 cos( t ) , (9 - x 2 ) 3 / 2 = 27 cos 3 ( t ), and Z dx (9 - x 2 ) 3 / 2 = Z 1 27 cos 3 ( t ) 3 cos( t ) dt = 1 9 Z sec 2 ( t ) dt = 1 9 tan( t )+ C = 1 9 x 9 - x 2 + C . x 3 y sin( t )= x 3 y = 9 - x 2 (see figure) tan( t )= x 9 - x 2 . 2. Evaluate the following limits: [4] (a) lim x →∞ 4 x 3 - 1 2 x 3 + x +3 (b) lim x 0 1 - cos(2 x ) 4 x 2 Solution: By L’H b o pital’s rule, (a) lim x →∞ 4 x 3 - 1 2 x 3 + x +3 = lim x →∞ 12 x 2 6 x 2 +1 = lim x →∞ 24 x 12 x = lim x →∞ 24 12 =2 . (b) lim x 0 1 - cos(2 x ) 4 x 2 = lim x 0 2 sin(2 x ) 8 x = lim x 0 4 cos(2 x ) 8 = 1 2 . 3. Evaluate Z 3 0 1 1+ x 2 dx . [6] Solution: Let x = tan( t ) , - π 2 <t< π 2 . Then 1+ x 2 = sec( t ) , dx dt = sec 2 ( t ) ,x =0 t =0 , x = 3 t = π 3 Z 3 0 1 1+ x 2 dx = Z π 3 0 sec( t ) dt =ln | sec( t ) + tan( t ) | π 3 0 = ln(2 + 3). 4. Write down the form of the partial fraction decomposition of x 4 +5 x 3 +1 ( x +1) 3 ( x - 3) 2 , but do not [2] determine the values of the constants A, B , etc. Solution: x 4 +5 x 3 +1 ( x +1) 3 ( x - 3) 2 = A x +1 + B ( x +1) 2 + C ( x +1) 3 + D x - 3 + E ( x - 3) 2

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2 5. Evaluate Z 5 x +1 ( x - 1)( x 2 +1) dx [6] Solution: 5 x +1 ( x - 1)( x 2 +1) = A x - 1 + Bx + C x 2 +1 = A ( x 2 +1)+( Bx + C )( x - 1) ( x - 1)( x 2 +1) A + B =0 , - B + C =5 ,A - C =1 A
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Unformatted text preview: = 3 , B =-3 , C = 2 ⇒ Z 5 x + 1 ( x-1)( x 2 + 1) dx = Z 3 x-1 +-3 x + 2 x 2 + 1 dx = Z 3 x-1 dx-3 2 Z 2 x x 2 + 1 dx + Z 2 x 2 + 1 dx = 3 ln | x-1 | -3 2 ln( x 2 + 1) + 2 tan-1 ( x ) + C . 6. For each of the following improper integrals, determine whether it converges or diverges. Eval-[6] uate the ones which converge. (a) Z ∞ 4 xe-x 2 dx (b) Z 1-1 1 x 3 dx (c) Z 1 1 √ x dx Solution: (a) Z ∞ 4 xe-x 2 dx =-2 e-x 2 ∞ = 2. The integral converges to 2. (b) Z 1-1 1 x 3 dx = Z-1 1 x 3 dx + Z 1 1 x 3 dx =-1 2 x 2-1-1 2 x 2 1 =-∞ + ∞ . The integral diverges. (c) Z 1 1 √ x dx = 2 √ x 1 = 2. The integral converges to 2....
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## This note was uploaded on 01/18/2012 for the course MATH 2007 taught by Professor S.melkonian during the Fall '11 term at Carleton CA.

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Solutions2 - = 3 , B =-3 , C = 2 ⇒ Z 5 x + 1 ( x-1)( x 2...

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