Exact area (1 Viewer)

[atakach99]

Find the exact area bounded by the curves y=sin x and y= 1/2 in the domain 0< x < 2pie

 

[tommykins]

This one was a huge bitch to do.

I suggest you draw a diagram, find points of intersection and split up the integral into separate parts.

ie. int 0 -> pi/4 cosx - sinx would be your first part (as the cosx graph is above the sin x graph)

 

[atakach99]

sorry my mistake its actually y=sin x and y=1/2

 

[tommykins]

In which domains?

Find points of intersection, split up the integral into 4 different integrals (judging from my graph and the domains 0 < x < 2pi) and add them all together.

Answer should be approximately 3.18 units²

 

[atakach99]

yea thats what i did

the answer at the back is root3 - pie /3 units squared

 

[tommykins]

I reckon this is a stupid question, it only takes into account the area when sin x is above the line y = 1/2

anyways, the integral is -

5pi/6
int sin x - pi/3
pi/6

This is because sin x is at the top of the line y = 1/2 in this domain. The area of the rectangle in this domain is given by [5pi/6 - pi/6]*1/2