Calculate the area of the plane region bounded by the graph of y=sin⁡x , the X-axis and the vertical line x=5π/2

From the fig. it is clear that sin⁡x=0  at x=0 ,π,2π and  sin⁡x≥0  in the intervals[0,π]   and  [2π,5π/2], but sin⁡x≤0   in the interval[π,2π], thus the required area is:
A=∫_a^b〖f(x)〗 dx=∫_0^(5π/2)|sin⁡x |  dx=∫_0^πsin⁡x  dx-∫_π^2πsin⁡x  dx+∫_2π^(5π/2)sin⁡x  dx
=-|cos⁡x |_0^π-[-cos⁡x ]_π^2π+[-cos⁡x ]_2π^(5π/2)
-(cos⁡π-cos⁡0 )-(-cos⁡2π+cos⁡π )+(-cos⁡〖5π/2〗+cos⁡2π )
=-(-1-1)-(-1-1)+(0+1)=5 square units.