CalculusSuperchallenge ShareCalculus · Superchallenge problemCompute ⌊1000⋅L⌋\left\lfloor 1000 \cdot L \right\rfloor⌊1000⋅L⌋ where L=∫011+x2 dxL = \displaystyle\int_0^{1} \sqrt{1+x^2}\,dxL=∫011+x2dx (the arc length of y=x22y = \tfrac{x^2}{2}y=2x2 from 000 to 111).Check Show answer Walk through it step by step Practice more like this