Areas between curves

Question

Find the area under y=3sin(x) and above 3cos(x) for pi/2 < x < 3pi/2

Answer 1

Let's say:

f(x) = 3•sin(x)

g(x) = 3•cos(x)

So if we are looking for an area under f(x) and above g(x) for π/2 < x < 3π/2, the definite integral will be:

∫^b_a [f(x) - g(x)] dx,

BE CAREFUL!: b ≠ 3π/2, because for π/2 < x < 5π/4, f(x) is above g(x), but for 5π/4 < x < 3π/2, g(x) is above f(x). So we choose b = 5π/4. Check it with any graphing utility.

Substitute:

∫^5π/4_π/2 [3•sin(x) - 3•cos(x)] dx

= [-3•cos(x) - 3•sin(x)]^5π/4_π/2

= -3•cos(5π/4) - 3•sin(5π/4) - [-3•cos(π/2) - 3•sin(π/2)]

= -3[-(2)^1/2/2] - 3 [-(2)^1/2/2] - [-3•(0) - 3•(1)]

= 6[(2)^1/2/2] + 3

= 3•(2)^1/2+ 3

≈ 7.24 sq. units

Answer 2

3cosx = 3sinx; tanx = 1; x = π/4, x = 5π/4;

Area A = ∫_π/2^5π/4(3sinx - 3cosx)dx = (- 3cox - 3sinx)_π/2^5π/4 = (3√2/2 + 3√2/2) - ( - 3) = 3√2 + 3