2011_Solution_3 - MA 1505 Mathematics I Tutorial 3...

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1505 Mathematics I Tutorial 3 Solutions 1. (a) Z 2 1 s 2 + s s 2 ds = Z 2 1 (1 + s - 3 / 2 ) ds = ( 2 - 1) - 2 s - 1 / 2 2 1 = ( 2 - 1) - 2 4 2 + 2 = 1 + 2 - 2 3 / 4 . (b) Z 4 - 4 | x | dx = Z 4 0 xdx + Z 0 - 4 ( - x ) dx = 1 2 4 2 + 1 2 4 2 = 16 . (c) Z π 0 1 2 (cos x + | cos x | ) dx = Z π/ 2 0 1 2 (cos x + | cos x | ) dx + Z π π/ 2 1 2 (cos x + | cos x | ) dx = Z π/ 2 0 cos xdx + 0 = sin x π/ 2 0 = 1 . (d) Z π 0 sin 2 1 + θ 2 · = Z π 0 1 2 £ 1 - cos(2 + θ ) / = 1 2 π - 1 2 sin(2 + θ ) π 0 = 1 2 π - 1 2 £ sin(2 + π ) - sin2 / = 1 2 π + sin2 . 2. The Fundamental Theorem of Calculus (I) says that d du Z u a f ( t ) dt = f ( u ) for a continuous function f . Here a is a fixed number. It is a sort of chain rule to find d dx Z g ( x ) a f ( t ) dt. To see this, let F ( u ) = Z u a f ( t ) dt and u = g ( x ) . It follows that dF du = d du Z u a f ( t ) dt = f ( u ) . Furthermore,
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This note was uploaded on 01/05/2012 for the course MATHEMATIC MA1505 taught by Professor Freudleong during the Spring '10 term at National University of Singapore.

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2011_Solution_3 - MA 1505 Mathematics I Tutorial 3...

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