find the area y = 3 cos(4x), y = 3 − 3 cos(4x), 0 ≤ x ≤ 𝜋/4

Answer 1

Let find x-coordinate of intersection points: 3 - 3cos(4x) = 3cos(4x); cos(4x) = 1/2; 4x = π/3 or x = 5π/3;

x = π/12 or x = 5π/12;so π/12 ≤ x ≤ π/4.

Area A = ∫₀^π/12(3cos(4x) - 3 + 3cos(4x))dx ₊ ∫_π/12^π/4 (3 - 3cos(4x) - 3cos(4x))dx = (3/2sin(4x) - 3x)₀^π/12 +

(3x - 3/2sin(4x))_π/12^π/4 = 3/2sin(π/3) - π/4 + 3π/4 - 3/2sin(π) - π/4 + 3/2sin(π/3) = π/4 + 3√3/2= (π + 6√3)/4 ≈

3.3835

Answer 2

f(x) = 3cos(4x) and g(x) = 3-3cos(4x)

If you plot the curves, you will see there are two areas for intervals [0,π/12] and [π/12, π/4]

π/12 π/4

∫ |f(x) - g(x)| dx + ∫ |g(x) - f(x)|dx

0 π/12

π/12 π/4

∫ |3cos(4x)-(3-3cos(4x)| dx + ∫ 3-3cos(4x) - 3cos(4x)|dx

0 π/12

[6sin(4x)/4 -3x]₀^π/12 + [3x -6sin(4x)/4]_π/12 ^π/4



(3/2)√3/2 + π/4 + (3/2)√3/2 = (π+6√3)/4