The document asks how to find the length of a curve and area under the curve defined by y=sin2x+2 from x=0 to x=2pi. It also asks what would change if y=sin10x+2. To solve part A, the function sin2x+2 always lies above the x-axis, so the area can be found by integrating sin2x from 0 to 2pi, which is 0. The area is then 2*(2pi-0)=4pi. For part B with y=sin10x+2, sketching the graph would show it oscillates more rapidly, so the length and area would both increase compared to y=sin2x

1. Can someone solve/explain this? I thought I was understanding this, but I think I've confused
myself.
A. Find the length of the curve and the area under the curve where x=0, x=2pi, y=0, y=sin2x+2.
B. What if y=sin10x +2? What is the pattern about the length and/or area? (Sketch the graph)
Solution
A. here the function sin(2x) 2 always lies above x-axis(y=0). So we can directly
integrate from x=0 to x=2pi. integral of sin(2x) in the given range is 0. So the area under the
curve is 2*(2pi-0)= 4*pi