Let the region R be bounded by f(x)=sin x, the x-axis and the line x = pi/2.

Question

Find the perimeter of the region R.

Answer 1

The total perimeter of the region R will be the length of the arc from 0 to π/2, the horizontal distance from 0 to π/2 and the vertical distance at sin(π/2) = 1.

The length of the arc is ∫√(((f '(x))² + 1)dx integrating from 0 to π/2.

f '(x) = cos(x)

∫√(cos²(x) + 1)dx

Unfortunately, this can't be integrated to an elementary function. Will have to use numerical integration.

Using Simpson rule for n = 4:

dx = (b-a)/n = π/8

F(x) from 0 to π/2 ≈ dx/3(f(0) + 4f(π/8) + 2f(π/4) + 4f(3π/8) +f(π/2)]

≈ π/24( √2 + 4√(cos²(π/8) + 1) + 2√(cos²(π/4) +1) + 4√(cos²(3π/8) + 1) + √(cos²(π/2) + 1))

≈ 1.9101

So total length would be 1.9101 + π/2 + 1 ≈ 4.48

Answer 2

R is bounded by arc of f(x) = sinx, at the bottom piece of x-axis from 0 to π/2, so length π/2;

and from therite piece of line x = π/2 from 0 to 1, length 1.

Length of arc is s = ∫₀^π/2(1 + cos²x)^1/2dx = 3.1412 (I use TI-84 to evaluate this integral)

So perimetr p = 3.1412+ 1 + π/2 = 5.712.