CalculusEasyApplications of Integration ShareApplications of Integration · Easy problemSet up (do not evaluate) the integral for the arc length of y=x2y=x^2y=x2 from x=0x=0x=0 to x=2x=2x=2.A∫021+4x2 dx\displaystyle\int_0^2\sqrt{1+4x^2}\,dx∫021+4x2dxB∫031+4x2 dx\displaystyle\int_0^{3}\sqrt{1+4x^2}\,dx∫031+4x2dxC∫041+4x2 dx\displaystyle\int_0^4\sqrt{1+4x^2}\,dx∫041+4x2dxD∫021+4x3 dx\displaystyle\int_0^2\sqrt{1+4x^3}\,dx∫021+4x3dxCheck Show answer Walk through it step by step Practice more like this Review this concept