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MAT1332W10A2S - MAT 1332 Assignment#2 Solutions[5 1 If you...

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MAT 1332, Assignment #2 Solutions 1. If you simply integrate, the answer is zero. So that obviously isn’t right. It’s best to sketch the two graphs. [5] See Figure 1. 0 -1 -0.5 0 0.5 1 B C A sin(2x) cos(2x) 5 π /8 π /8 π /2 π /4 3 π /4 π Figure 1: Area between sin(2 x ) and cos(2 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 find them, we need to find the points of intersection. That is, the values of x where sin(2 x ) = cos(2 x ). If you look at Figure 1 again, you may be able to subtly discern these. But let’s figure them out: sin(2 x ) = cos(2 x ) tan(2 x ) = 1 (dividing both sides by cos(2 x )) 2 x = ... - 7 π 4 , - 3 π 4 , π 4 , 5 π 4 , 9 π 4 ... x = ... - 7 π 8 , - 3 π 8 , π 8 , 5 π 8 , 9 π 8 ... Actually, since 0 x π , we only need values that fall into this range. So the points of intersection within this range are x = π 8 and x = 5 π 8 . Now we’re ready to integrate. The area is the sum of areas A, B and C, so we have Area = A + B + C = Z π/ 8 0 (cos(2 x ) - sin(2 x )) dx + Z 5 π/ 8 π/ 8 (sin(2 x ) - cos(2
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