{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MAT1332W10A2S

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}