How to find area under the curve. Sin(x) from 0 to 2Pi. Explain in details

√( 1 + (dy/dx)^2 ) √( 1 + (cos(x))^2 ) √( 1 + cos^2(x) ) 2π ∫ √( 1 + cos^2(x) ) dx ≈ 7.64 0 Area: 2π ∫ (sin(x) + 2 - 0) dx 0 2π ∫ sin(x) + 2 dx 0 2π - cos(x) + 2x ] 0 - ( cos(2π) - cos(0) ) + 2 * (2π - 0) - ( 1 - 1 ) + 2 * (2π) 4π

That is the length of the curve, but that integral has no closed form, however, it does result in a real number, so technically it's a real number and thus a length.

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