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July 30, 2014

July 30, 2014

Posted by **lssa** on Thursday, April 8, 2010 at 4:19pm.

y=10cosx, y=10sin2x,x=0

- calculus -
**Reiny**, Thursday, April 8, 2010 at 4:53pmfirst solve the two to find where they cross

10sin2x = 10cosx

sin2x - cosx = 0

2sinxcox - cosx = 0

cosx(2sinx - 1) = 0

cosx = 0 or sinx = 1/2

x = π/2 or x = π/4 in the first quadrant

so I would do it in two integrals

1. integral [ 10sin2x] from 0 to π/4

2. integral [10cosx] from π/4 to π/2

Area = -20cos2x | from 0 to π/4 + 10sinx | from π/4 to π/2

= (-20cos2(π/4) + 20cos 0) + (10sinπ/2 - (10sinπ/4))

= 0 + 20 + 0 - 10/√2

= 20 - 10/√2

= (20√2 - 10)/√2 = appr. 12.929

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