1007_Tut5_Sol

# 1007_Tut5_Sol - du = 1 x dx v =-1 x Z u dv = u v-Z v du Z...

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MATH 1007 A Tutorial #5 Nov 24, 2006 SOLUTIONS 1. [4 marks] Find the area between y = 3 x 2 and y = 3 x + 6. (Show algebraically where these 2 functions meet, and give a rough sketch of the functions in your answer.) Solution: First we determine where these 2 functions meet: 3 x 2 = 3 x +6 = 3 x 2 - 3 x - 6 = 0 = 3( x 2 - x - 2) = 0 = 3( x - 2)( x +1) = 0 = x = +2 , x = - 1 Hence y = 3 x 2 and y = 3 x + 6 meet when x = +2 , x = - 1. Note that y (2) = 12 and y ( - 1) = 3. Area = Z 2 - 1 top - bottom dx = Z 2 - 1 (3 x + 6) - (3 x 2 ) dx = ± 3 2 x 2 + 6 x - x 3 ² 2 - 1 = ³ 3 · 4 2 + 12 - 8 ´ - ³ 3 2 - 6 + 1 ´ = 6 + 12 - 8 - 3 2 + 6 - 1 = 24 - 9 - 3 2 = 15 - 3 2 = 13 1 2 1

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2. [3 marks] Evaluate Z ln x x 2 dx Solution: Let u = ln x dv = 1 x 2 dx then
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Unformatted text preview: du = 1 x dx v =-1 x Z u dv = u v-Z v du Z ln x x 2 dx = (ln x ) ±-1 x ²-Z ±-1 x ²± 1 x ² dx =-ln x x + Z 1 x 2 dx =-ln x x-1 x + C 3. [3 marks] Evaluate Z 1 x 2 (1 + 2 x 3 ) 5 dx . (Hint: 3 6 = 729) Solution: Set u = 1 + 2 x 3 x = 0 = ⇒ u = 1 Then du = 6 x 2 dx x = 1 = ⇒ u = 3 Z 1 x 2 (1 + 2 x 3 ) 5 dx = Z 1 1 6 (1 + 2 x 3 ) 5 6 x 2 dx = Z 3 1 1 6 u 5 du = ³ 1 36 u 6 ´ 3 1 = 3 6 36-1 36 = 729-1 36 = 728 36 2...
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1007_Tut5_Sol - du = 1 x dx v =-1 x Z u dv = u v-Z v du Z...

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