Find the area under one arch of curve y=a sin x/a). This lesson is plane areas in rectangular coordinates. I don't know how to solve it. Thanks for your help.
The way you typed it, the a's would cancel out, and your equation would
simply be y = sinx
So you must have meant : y = asin(x/a)
(see how critical those brackets are )
for one "arch" , the x-intercepts are x/a = 0 and x/a = π
x = 0 to x = aπ
area = ∫ asin(x/a) dx from 0 to aπ
= [-acos(x/a)(a) ] from 0 to aπ
= [ - (a^2) cos(x/a) ] from 0 to aπ
= -a^2 cos(π) - (-a^2 cos 0)
= a^2 + a^2
= 2a^2