1007_Tut4_Sol

# 1007_Tut4_Sol - MATH 1007 A Tutorial #4 Nov 10, 2006...

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MATH 1007 A Tutorial #4 Nov 10, 2006 Solutions 1. [4 marks] Use substititution to ﬁnd the required integrals: (a) Z (4 + x 2 ) 10 (2 x ) dx Hint: Set u = 4 + x 2 Solution: Then du dx = 2 x = du = 2 x dx Z (4 + x 2 ) 10 (2 x ) dx = Z u 10 du = 1 11 u 11 + C = 1 11 (4 + x 2 ) 11 + C (b) Z 2 x 4 - 1 (8 x 3 ) dx Hint: Set u = 2 x 4 - 1 Solution: Then du dx = 8 x 3 = du = 8 x 3 dx Z 2 x 4 - 1 (8 x 3 ) dx = Z u du = Z u 1 / 2 du = 1 3 / 2 u 3 / 2 + C = 2 3 (2 x 4 - 1) 3 / 2 + C 1

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2. [4 marks] Use the Fundamental Theorem of Calculus to solve the fol- lowing questions: (a) Z 4 1 (1 + 6 x ) dx Solution: Z 4 1 (1 + 6 x ) dx = ± x + 3 x 2 ² 4 1 = (4 + 3 · 16) - (1 + 3 · 1) = (4 + 48) - 4 = 48 (b) If g ( x ) = Z e x 0 sin 3 t dt , ﬁnd g 0 ( x ). Hint: let u = e x so that dg dx = dg du · du dx Solution: dg dx = dg du · du dx = d du ³Z u 0 sin 3 t dt ´ · d dx ( e x ) = (sin 3 u )( e x ) = e x sin 3 ( e x ) 2
3. [2 marks] Consider the function f ( x ) = 1 x from x = 1 to x = 5. Use 4 approximating rectangles/intervals to give an expression for the area under f (a) using right end points of intervals (b) using left end points of intervals
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## This note was uploaded on 03/22/2010 for the course MATH 1007 taught by Professor Unknown during the Fall '07 term at Carleton.

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1007_Tut4_Sol - MATH 1007 A Tutorial #4 Nov 10, 2006...

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