# calculus

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Sketch the region in the first quadrant enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=10cosx, y=10sin2x,x=0

• calculus -

first 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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