find the area of the regions enclosed by the given curves. y = 7 cos 5x, y = 7 s

Question

find the area of the regions enclosed by the given curves. y = 7 cos 5x, y = 7 sin 10x, x = 0, x = ?/10

Explanation / Answer

First, find the points of intersection by equating the y values. 4cos(5x) = 4sin(10x) ==> 4cos(5x) = 8sin(5x)cos(5x), by the double-angle formula for sine ==> 8sin(5x)cos(5x) - 4cos(5x) = 0 ==> 4cos(5x)[2sin(5x) - 1] = 0 ==> cos(5x) = 0 and sin(5x) = 1/2. This solves to get: 5x = p/2 ± 2pk, 5x = 3p/2 ± 2pk, 5x = p/6 ± 2pk, and 5x = 5p/6 ± 2pk ==> x = p/10 ± 2pk/5, x = 3p/10 ± 2pk/5, x = p/30 ± 2pk/5, and x = p/6 ± 2pk/5. Om the interval, x = p/30 and x = p/10 are your solutions. Since 4cos(5x) > 4sin(10x) on (0, p/30) and 4cos(5x) < 4sin(10x) on (p/30, p/10), the area is: ? [4cos(5x) - 4sin(10x)] dx (from x=0 to p/30) + ? [4sin(10x) - 4cos(5x)] dx (from x=p/30 to p/10). Upon integrating, the area equals 2/5 square units. I hope this helps!

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