Find the area of the region.

Question

Sketch the region enclosed by the given curves. Find its area.

y = tan(7x),    y = 2 sin(7x),    −𝜋/21 ≤ x ≤ 𝜋/21

Answer 1

Area between the two curves could refer to

∫(tan(7x) - 2sin(7x))dx or ∫|tan(7x) - 2sin(7x)|dx

Either way we need to see where the difference between the two functions

is positive and where it is negative.

tan(7x) - 2sin(7x) =

sin(7x)/cos(7x) - 2sin(7x) =

sin(7x)(1/cos(7x) - 2)

Given

-π/21 ≤ x ≤ π/21

-π/3 ≤ 7x ≤ π/3

cos(7x) ≥ 1/2

1/cos(7x) ≤ 2

1/cos(7x) -2 ≤ 0

When 0 ≤ 7x ≤ π/3 then sin(7x) ≥ 0,

When -π/3 ≤ 7x ≤ 0 then sin(7x) ≤ 0.

Combining the above results,

when 0 ≤ x ≤ π/21 then tan(7x) ≤ 2sin(7x),

when -π/21 ≤ x ≤ 0 then tan(7x) ≥ 2sin(7x)

Furthermore, on the interval (-π/21, π/21) we have the symmetry

sin(7x) = -sin(-7x)

tax(7x) = -tax(-7x)

Therefore

∫_-π/21⁰ (tan(7x) - 2sin(7x)) dx = -∫₀^π/21 (tan(7x) - 2sin(7x)) dx

Therefore

∫_-π/21^π/21 (tan(7x) - 2sin(7x)) dx = 0

and

∫_-π/21^π/21 |tan(7x) - 2sin(7x)| dx = 2∫₀^π/21 (2sin(7x) - tan(7x)) dx

To evaluate the latter, first do the indefinite integrals

∫2sin(7x)dx = (-2/7)cos(7x)

∫tan(7x)dx = (-1/7)ln(cos(7x))

The final result is then this definite integral

2∫₀^π/21(2sin(7x) - tan(7x))dx =

2((-2/7)cos(7×π/21) - (-2/7)cos(0)) - 2((-1/7)ln(cos(7×π/21)) - (-1/7)ln(cos(0))) = 0.088