area beneath the curve. help plz. (1 Viewer)

[Dio-Rama]

find the area beneath the curve y=sinx,
between the x-axis and the domain
0 (lessthan,equal to) x (less than, equal to) 2pi.

 

[SoulSearcher]

Split the integral into 2 sections, since from pi < x < 2pi, the curve is below the x-axis.

int. [sin x](2pi->0)
= int [sin x](pi->0) + l int [sin x](2pi->pi) l
= [-cos x](pi->0) + l [-cos x](2pi->pi) l
= (-cos pi) - (-cos 0) + l (-cos 2pi) - (-cos pi) l
= 1 + 1 + l(-1-1)l
= 2 + 2
= 4 units²

 

[bananasmoothy]

Uses less ink, paper and, more importantly, thinking if you do:
4 X int. from 0->pi/2 (sin x) dx
= 4 [-cos x] (0 -> pi/2)
= 4 [-cos pi/2 - - cos 0] (draw mini cos curve)
= 4 (0 + 1)
= 4 units sq.

 

[SoulSearcher]

Works both ways. Also possible to do it by multiplying the integral of y = sin x with the limits 0 and pi by 2, i.e. 2 * int. (0->pi)[sin x] dx.