Calculating Area Under a Sine Curve on a Limited Interval

[naaa00]

Homework Statement

Find the area of: {(x, y) R^2 : 0 ≤ x ≤ π, 0 ≤ y ≤ 1/2, 0 ≤ y ≤ sin x}

The Attempt at a Solution

y = sinx on the interval 0 to π but with y < 1/2.

if Sinx = 1/2, then x = pi/6, or x = 5pi/6

Because of symmetry I integrate one part and the multiply by two.

[0,pi/6) In: (sinx) dx = (-cosx) = -cos(π/6) - -cos(0) = cos(0) - cos(π/6) = 1 - (1/2)√3.

2(1 - (1/2)√3.)

I stop there, but then I saw that to the answer that I had, I must add the rectangular area from pi/6 to 1/2(pi) with height 1/2, or (1/3)pi(1/2) = pi/6

So the answer was: 2(1 - (1/2)√3 + pi/6) = 2 - √3 + (1/3)pi).

I don't understand that last part. Which rectangular area? I don't see it.

 

[ehild]

I don't understand that last part. Which rectangular area? I don't see it.

Have you plotted the function and the boundaries?

ehild

 

[naaa00]

lol, ok. Now I saw the rectangular area...

Thank you!