Assignment 2 - MAT 1332, Winter 2007, Assignment #2,...

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MAT 1332, Winter 2007, Assignment #2, Solutions Total marks=11. 1. If you simply integrate, the answer is zero. So that obviously isn’t right. It’s best to sketch the two graphs. [2] See Figure 1. 0 ± 1 ± 0.5 0 0.5 1 B C A ± 3 ± /16 cos(4x) ± 7 ± /16 ± ± /2 sin(4x) Figure 1: Area between sin(4 x ) and cos(4 x ). Clearly the area between them isn’t zero. In fact, it’s area A, plus area B, plus area C on the graph. And before we can ±nd them, we need to ±nd the points of intersection. That is, the values of x where sin(4 x ) = cos(4 x ). If you look at Figure 1 again, you may be able to subtly discern these. But let’s ±gure them out: sin(4 x ) = cos(4 x ) tan(4 x ) = 1 (dividing both sides by cos(4 x )) 4 x = ... 7 π 4 , 3 π 4 , π 4 , 5 π 4 , 9 π 4 ... x = ... 7 π 16 , 3 π 16 , π 16 , 5 π 16 , 9 π 16 ... Actually, since π 2 x 0, we only need values that fall into this range. So the points of intersection within this range are x = 7 π 16 and x = 3 π 16 . Now we’re ready to integrate. The area is the sum of areas A, B and C, so we have Area = A + B + C = ! 7 π/ 16 2 (cos(4 x ) sin(4 x )) dx + ! 3 16 7 16 (sin(4 x ) cos(4 x )) dx + ! 0 3 16 (cos(4 x ) sin(4 x )) dx 1
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= ± sin(4 x ) 4 + cos(4 x ) 4 ² 7 π/ 16 2 + ± cos(4 x ) 4 sin(4 x ) 4 ² 3 16 7 16 + ± sin(4 x ) 4 + cos(4 x ) 4 ² 0 3 16 = ±³ sin( 7 4) 4 + cos( 7 4) 4 ´ ³ sin( 2 π ) 4 + cos( 2 π ) 4 ´² + ±³ cos( 3 4) 4 sin( 3 4) 4 ´ ³ cos( 7 4) 4 sin( 7 4) 4 ´² + µ ³ sin(0) 4 + cos(0) 4 ´ sin( 3 4) 4 + cos( 3 π 4 ) 4 ·¸ = ±³ 1 4 2 + 1 4 2 ´ ³ 0+ 1 4 ´² + ±³ 1 4 2 + 1 4 2 ´
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This note was uploaded on 02/23/2012 for the course MAT MAT1332 taught by Professor Arianemasuda during the Spring '09 term at University of Ottawa.

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Assignment 2 - MAT 1332, Winter 2007, Assignment #2,...

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